"Summer letters" of the year 2000 (English translation, April 2023)
(towards the foundations of the semi-infinite cohomology theory for associative algebras)
Russian original is here -- https://posic.livejournal.com/413.html
The continuation (two Summer 2002 letters) is here -- https://posic.livejournal.com/2774248.html
A series of letters about the semi-infinite (co)homology of
associative algebras
This is my April 2023 English translation of my Summer 2000 and
Summer 2002 letters to Roma Bezrukavnikov and Serezha Arkhipov
with Spring 2006 additions/comments.
This translation keeps most peculiarities of the terminology,
assumptions etc. of the original 2000-02 letters. The reader
will notice the usage of "free/cofree module" assumptions where
"projective/injective" would certainly suffice, etc. The notation
like A-mod stands for the category of all A-modules, not just
finitely generated or finite-dimensional ones.
One exception is that a particularly conflicting and outdated
early terminology of "comodule algebra" is replaced by my modern
term "semialgebra" (to avoid creating any further confusion).
The reader is also advised to keep in mind that what was called
"the derived category D-prime" in early 2000's is now known as
the coderived category, while what was then called "the derived
category D-second" is now known as the contraderived category.
Several misprints found in the Russian original have been
corrected in this translation.
From: Leonid Positselski < posic@mccme.ru >
To: roma, hippie
Date: Fri, 1 Sep 2000 09:40:45 +0400 (MSD)
Subject: demystifying A^#, first part (with corrections)
Lines: 65
Hi Roma and Serezha,
I have made some further corrections beyond those already mentioned,
all concerning the left/right issues, and also some additions.
Thanks for your attention etc.
I. Description of the algebra A^# in the case when A=N\otimes B,
where N and B are subalgebras. Uses the definition of A^# that
Serezha usually gives. Clarifies the infinite-dimensional
situation a little bit. Otherwise it is useless.
Let A be a graded algebra with two graded subalgebras N and B
such that N_i is finite-dimensional for all i, N_i=0 for i<<0,
and B_i=0 for i>>0. Assume that the multiplication map
N\otimes_k B \to A is an isomorphism. Then the algebra A can be
uniquely recovered from N, B and the "transposition" map
\phi: B\otimes N \to N\otimes B.
By "raising the indices" one can construct from the map \psi a map
\psi: N^*\otimes B \to Hom_k(N,B) = B\otimes N^*
(the latter equality follows from the conditions on the gradings
of N and B. Hom_k denotes the graded Hom space, i.e., the direct
sum of the spaces of homogeneous maps of various degrees).
Now I digress to recall Serezha's definition of the modules S
and S'. So S=N^*\otimes_N A and S'=Hom_{B-right}(A,B) (the right
B-module Hom). These are two right A-modules. There is a map
S \to S' between them,
n^*\otimes_N a \mapsto (a'\mapsto < n^*, aa' >)
where < n^*, - >: A \to B is the map pairing n^* with the first
tensor factor in A=N\otimes B.
We have S = N^*\otimes_k B = S' as vector spaces. I claim that
the A-module map S \to S' described above coincides with
the vector space map \psi described above. For Serezha's
definition of the algebra A^# to make sense, it is needed that
\psi be an isomorphism. I am assuming that.
Now consider the inverse map
\psi^{-1}: B\otimes N^* \to N^*\otimes B.
I want to "lower its indices" back. Then one obtains a certain map
\phi^#: N\otimes B \to Hom_k(N^*,B).
Unfortunately, now it does not follow from anything that the image
of \phi^# is contained in the subspace of maps N^*\to B of finite
rank, i.e., in B\otimes N \subset Hom_k(N^*,B). I claim that
Serezha's algebra A^# exists precisely when this is so, and in this
case the algebra A^# is contracted from the "transposition map"
\phi^#.
In the finite-dimensional case the latter condition is trivial, so
it is sufficient that the map \psi be invertible.
Now we can define the algebra A^# without using the modules S and S',
but using only the maps \phi and \phi^#. Specifically, for the
existence of an algebra with a given "transposition" map it is
necessary and sufficient that such map satisfies certain identities
(associativity). One can check that \phi and \phi^# satisfy these
identities simultaneously.
Lenya.
From: Leonid Positselski < posic@mccme.ru >
To: roma, hippie
Date: Fri, 1 Sep 2000 09:43:12 +0400 (MSD)
Subject: demystifying A^#, second part (with corrections)
Lines: 77
II. Description of the algebra A^# in the finite-dimensional case.
Uses neither the subalgebra B nor the grading. Only uses
the subalgebra N. I recommend it.
Let us start with a digression. Let N be a finite-dimensional
algebra (literally finite-dimensional; a graded locally
finite-dimensional would not do). Then N^* is a coalgebra, and
N-modules are the same things as N^*-comodules. On N-modules,
there is the operation of tensor product, while on N^*-comodules
there is the cotensor product. (If one wishes, the latter can be
defined in module terms as Hom_{N-bimod-N}(N,P\otimes_k Q),
where P and Q are a right and a left N-module, respectively.)
We will denote the tensor product of N-modules by P\ot_N Q and
the cotensor product by P\oc_{N^*} Q. On the category of
N-modules, there are (Serezha's) endofunctors P \mapsto N^*\ot_N P
and P \mapsto N\oc_{N^*}P = Hom_N(N^*,P).
I claim that for any right N-module P and left N-module Q there is
a map (of abelian groups)
(1) P\ot_N Q \to P\oc_{N^*} (N^*\ot_N Q),
which is an isomorphism, at least, when the module Q is free or
the module P is cofree. Analogously, there is a map
(2) P\ot_N (N\oc_{N^*} Q) \to P\oc_{N^*} Q,
and it is an isomorphism if the module Q is cofree or the module P
is free. (A kind of mutual associativity of the tensor and cotensor
products. A simple exercise.)
End of digression.
Now consider the tensor category of N-bimodules (with respect to
the tensor product). Bimodules that are free (say) on the left
form a tensor subcategory there. Analogously, there is the tensor
category of N-bimodules with respect to the cotensor product.
The left-cofree bimodules form a tensor subcategory in it. Then
the "associativity" formulas (1-2) mean that:
(i) the functor Q \mapsto N^*\ot_N Q is a tensor functor from
left-free bimodules w.r.t. the tensor product to left-cofree
bimodules w.r.t. the cotensor product;
(ii) Q \mapsto N\oc_{N^*} Q is a tensor functor in the opposite
direction;
(iii) these two functors are mutually inverse, and consequently,
equivalences of tensor categories.
In particular, these functors induce equivalences of the categories
of algebra objects in the first and the second tensor category.
Now let A be a (usual) algebra containing a subalgebra N; then A
is also an algebra in the category of N-bimodules. We see that if
A is free over N on the left, then S = N^*\ot_N A is a semialgebra
over N (i.e., an algebra with respect to the cotensor product).
Furthermore, if S turns out to be cofree on the right (and not only
on the left) over N, then one can define
A^# = S \oc_{N^*} N = (N^* \ot_N A) \oc_{N^*} N.
This is an algebra in the category of N-bimodules with respect to
the tensor product, i.e., simply an associative algebra containing N.
The condition that S is a cofree right N-module corresponds to
the condition of invertibility of the map \psi from my first letter.
The same formulas (1-2) imply that
(iv) right A-modules are the same things as right S-modules
(i.e. right N-modules with an S-module structure w.r.t. the cotensor
product);
(v) left A^#-modules are the same things as left S-modules;
(vi) the category of N-free left A-modules is equivalent to
the category of N-cofree left S-modules;
(vii) the category of N-free right A^#-modules is equivalent to
the category of N-cofree right S-modules.
Lenya.
From: Leonid Positselski < posic@mccme.ru >
To: roma, hippie
Date: Fri, 1 Sep 2000 09:59:50 +0400 (MSD)
Subject: demystifying A^#, third part (with corrections and additions)
Lines: 120
III. The semialgebra S and modules over it. A description using
neither the subalgebra B nor the grading. The infinite-dimensional
case.
As usual, I start with a digression. Let N be an algebra and C
a coalgebra (over the same field). I recall that there is a natural
algebra structure on the vector space C^* (such that any C-comodule
is a C^*-module). Suppose given an algebra homomorphism f: N\to C^*
with a dense image. In other words, there should be given a pairing
C\otimes N \to k that is compatible with the algebra and coalgebra
structures and nondegenerate in the second argument.
The datum of N, C and f is equivalent to the datum of a full
subcategory in the category of (say, left) N-modules with
the following properties: it is closed under subobjects and quotients
(but not necessarily under extensions!) and every module from this
subcategory is the union of its finite-dimensional submodules.
Specifically, to a triple (N, C, f) the fully faithful functor of
"restriction of scalars" f^*: C-comod \to N-mod is assigned.
A typical example: if H is an algebraic group (over a field of
characteristic 0), then there is the associated triple (N, C, f),
where N=U(h), h is the Lie algebra of H, and C=C(H) is the coalgebra
of functions on H. This triple corresponds to the full subcategory
in U(h)-mod consisting of the representations that can be integrated
to H.
Another example: if N is a graded algebra such that all N_i are
finite-dimensional and N_i=0 for i<<0, then one can take C to be
the graded dual vector space N^* to N. Then the N^*-comodules are
the N-modules with the property that for every element x from
such module one has N_i x = 0 for all i>>0.
A banal example: if N is finite-dimensional, one can take C=N^*.
Then all N-modules are C-comodules.
End of digression.
So, suppose given a triple (N, C, f). I will presume that
the categories of (left and right) C-comodules are embedded into
the respective categories of N-modules by means of f^*.
Accordingly, a statement like "such-and-such right N-module
is a C-comodule" means that this module belongs to the subcategory
of C-comodules in right N-modules.
Now in the tensor category of N-bimodules there is the following
tensor subcategory: a bimodule E belongs to this subcategory if for
any right C-comodule M the right N-module M\ot_N E is a C-comodule.
(It suffices that the N-module C\ot_N E be a right C-comodule.)
There is also the full subcategory, already familiar to us, of
N-bimodules that are free over N on the left. I am interested in
the intersection of these two full subcategories, i.e.,
the bimodules satisfying both the assumptions. I will call such
bimodules left (N,C,f)-admissible.
I claim that the functor E \mapsto C\ot_N E is a tensor functor
from left (N,C,f)-admissible bimodules to the bicomodules over C
(with respect to the cotensor product). So if N is a subalgebra
in A and A is left (N,C,f)-admissible, then S = C\ot_N A is
an algebra in the category of bicomodules over C. Moreover, let
M be a right C-comodule. Then to define a right A-module structure
on M (extending the right N-module structure related to the given
C-comodule structure) -- is the same thing as to define on M
a structure of right module, with respect to the cotensor product,
over the "semialgebra" S.
In other words, the full subcategory in the category of right
A-modules consisting of those modules which, viewed as N-modules,
are C-comodules, is isomorphic to the category of right modules
over S. This is some kind of infinite-dimensional generalization
of the property (iv) from the previous letter.
Both the claims are corollaries of the following "associativity"
for the tensor and cotensor products, generalizing formula (1): for
any right C-comodule P and left N-module Q there is a natural map
P \ot_N Q \to P\oc_C (C\ot_N Q),
which is an isomorphism if Q is free or P is cofree.
In fact, more generally, for any (unrelated) algebra N and
coalgebra C, comodule P and module Q, and a vector space T with
commuting structures of a left C-comodule and right N-module
there is a natural map
(P\oc_C T) \ot_N Q \to P \oc_C (T\ot_N Q),
which is an isomorphism if Q is free or P is cofree (while
no conditions are imposed on T).
Thus, as arguments of the functor of semi-infinite Tor one should
use:
- instead of bounded above graded right A-modules, while
the algebra N is positively graded, as Serezha had it --
right modules over the "semialgebra" S = C\ot_N A;
- instead of bounded above graded left A^#-modulej-- left
modules over S [see the previous letter, property (v)].
The final part will follow.
Lenya.
P.S. Remark: As in the previous letter, it is important here
that the semialgebra S should be C-cofree not only on the left
(which holds by construction), but also on the right. This is
needed, in particular, so that the category of left modules over S
were abelian (see also item 3 at the end of the fourth letter).
Suppose A = N\ot_k V as a left N-module, where V is some vector
space. Then S = C\ot_N A = C\ot_k V. The structure of right
C-comodule on S is given by a map S \to S\ot_k C. Denote by \psi_V
the composition of maps
C\ot_k V = S \to S\ot_k C = C\ot_k V\ot_k C \to V\ot_k C,
where the last arrow is obtained by applying the counit to the first
factor. As one can easily see, this is some kind of generalization
of the map \psi from the first letter. For S to be C-cofree on
on the right, it is obviously sufficient that there were a subspace
V \subset A for which \psi_V is invertible.
From: Leonid Positselski < posic@mccme.ru >
To: roma, hippie
Date: Fri, 1 Sep 2000 10:03:19 +0400 (MSD)
Subject: demystifying A^#, final part (with corrections and additions)
Lines: 364
IV. Contramodules. A description of the category where the second
argument of the semi-infinite Ext functor ranges. Nontrivial already
in the finite-dimensional case; works in the infinite-dimensional
case, too.
I start this final installment of these notes with a long
introduction (in an attempt to explain the problem to be solved).
In a reasonable sense, in the infinite-dimensional situation
the algebra A (or A^#) contains more information than
the semialgebra S. To construct A^# starting from S, as it was
explained in the first letter, one needs to have a subalgebra B.
If B is not given, one can only recover from S certain completions
of A and A^#: specifically,
A~ = End_{S-left}(S)^op and A^{#~} = End_{S-right}(S)
i.e., the endomorphisms of S as a left or right S-module. Indeed,
End_{S-right}(S) = End_{A-right}(S) = Hom_{N-right}(C, C\ot_N A)
= Hom_{C-right}(C, B\ot_k C) = Hom_k(C,B)
as a vector space (the first equality holds in view of
the equivalence of module categories from my previous letter,
the third one holds by virtue of a suitable version of
the isomorphism \psi).
In other words, it seems that from the semialgebra S one cannot
recover the whole category of right A-modules, but only
the "category O" -- the subcategory of those modules that are
C-comodules. The completion of the algebra A described above is
precisely the completion that acts on the modules from
"category O". (In the finite-dimensional case there are no
such distinctions, of course.)
I recall that (as Serezha taught us) the arguments of
the semi-infinite Ext functor are:
- The first argument ("from") -- left A^#-modules with
the grading bounded above. In the language of the semialgebra S,
these are left S-modules.
- The second argument ("to") -- left A-modules bounded below.
If the algebra N is infinite-dimensional, such modules are not even
C-comodules. So far now we did not know how to describe such
N-modules in terms of the coalgebra C, let alone describing
such A-modules in terms of the semialgebra S.
It will be explained below how to construct the abelian category
to which the second argument of the semi-infinite Ext functor
belongs in terms of the semialgebra S. This question is
surprizingly nontrivial in the finite-dimensional case already,
as -- following the assertions (iv-vii) from the second letter --
so far we could not recover the category of *left* A-modules (or
vice versa, right A^#-modules) in terms of S. We only know how to
recover the categories of right A-modules and left A^#-modules,
and also *N-free* left A-modules and right A^#-modules. Of course,
we want to use rather arbitrary modules and not just N-free ones
as arguments of the semi-infinite Ext. The abelian category of
"contramodules" which I will construct for a semialgebra S, will
coincide with the category of left A-modules in
the finite-dimensional case.
Surprizingly, one can even generalize the equivalences of
categories of N-free and N-cofree modules [statements (vi-vii) from
the second letter] to the infinite-dimensional case; so Roma's
definition of semi-infinite Ext can be likely generalized, too.
So, I start with a digression.
There are modules, there are comodules, and there are also
contramodules. For an infinite-dimensional coalgebra C one can
construct, alongside with the categories of left and right
comodules, two other abelian categories -- of left and right
contramodules. This is done in the following way.
Vector spaces (over a fixed field k) form a tensor catgory.
Therefore, one can consider algebras or coalgebras in this tensor
category; these are the usual algebras and coalgebras. Now if
one is interested in modules or comodules over such algebras or
coalgebras, one can define such (co)modules as objects of the same
category Vect. These will be the usual modules or comodules.
But one can also choose some *module category* over the tensor
category Vect, that is, a category M together with an "action"
functor Vect\times M \to M, with the suitable associativity
constraint -- and consider objects of this module category endowed
with a (co)module structure over a fixed (co)algebra from Vect.
On the category Vect^op opposite to Vect there is a structure of
(left) module category over Vect given by the formula
(V, W^op) \mapsto Hom_k(V,W)^op.
This is a module category in view of the isomorphism
Hom(U\ot_k V, W) = Hom(U, Hom(V,W)).
To obtain a right module category structure, one needs to use
another isomorphism
Hom(U\ot_k V, W) = Hom(V, Hom(U,W)).
in the role of the associativity constraint.
Now if B is an algebra in Vect, then a B-module in the module
category Vect^op over Vect is a called a contramodule over B.
More precisely, by a left B-contramodule one means an object of
the category opposite to the category of modules over B in
the *right* module category Vect^op (so that the forgetful functor
from contramodules to Vect is covariant). Simply put, a contramodule
is a vector space V together with a map V \to Hom_k(B,V), which must
satisfy certain associativity and unitality conditions. However,
to specify a map V \to Hom_k(B,V) is the same a to specify a map
B\ot_k V \to V; hence left B-contramodules are the same things as
left B-modules.
For coalgebras, the situation is different. The definition is
the same: if C is a coalgebra in Vect, then a left contramodule
over C is an object of the category opposite to the category of
comodules over C in the right module category Vect^op over Vect.
In other words, a contramodule over C is a vector space V together
with a map
Hom_k(C,V) \to V,
satisfying conditions of coassociativity and counitality of
the following form: two maps from the vector space
Hom_k(C, Hom_k(C,V)) = Hom_k(C\ot C, V)
to the space V -- one obtained by iterating the "contraaction"
operation and the other one depending on the comultiplication on C
-- must coincide; the composition of maps V \to Hom_k(C,V) \to V,
where the first arrow comes from the counit on C and the second one
is the "contraaction", must be the identity map.
On any C-contramodule (just as on any C-comodule) there is
a natural structure of a module over the algebra C^*, since
the space C^*\ot_k V is embedded into Hom_k(C,V). All the three
categories (left C-comodules, left C-contramodules, and left
C^*-modules) agree if the coalgebra C is finite-dimensional;
otherwise, they are all different. (In particular, even though
the operations of infinite direct sum and infinite product exist
on all the three categories, the functor from C-comodules to
C^*-modules only preserves the infinite direct sums, while
the functor from C-contramodules to C^*-modules only preserves
the infinite products, but not vice versa.)
Examples: (1) If M is a right comodule over C and U is an arbitrary
vector space, then the space Hom_k(M,U) has a natural structure of
left contramodule over C. More generally, if there is a structure
of left D-(co)module on M commuting with the structure of right
C-comodule, and U is some D-(co)module, then Hom_D(M,U) is a left
contramodule over C.
(2) Let N be a graded algebra with finite-dimensional components
such that N_i=0 dlya i<<0, and let N^* be the graded dual coalgebra.
Then if M is a graded left N-module bounded below, then on
the direct product of the grading components of the module M there
is a natural structure of left contramodule over N^*.
Contramodules of the form Hom_k(C,U), where the coalgebra C is
viewed as a right comodule over itself and U is an arbitrary vector
space, are called free C-contramodules. A short notation:
Hom_k(C,U) =: U^C. Exercise: there is a natural isomorphism
(a) Hom_{C-contra}(U^C, P) = Hom_k(U,P)
for any left C-contramodule P.
The next exercise I cannot even solve myself: is it true that
finite-dimensional contramodules over an infinite-dimenional
coalgebra C are the same things as finite-dimensional comodules;
that is, in other words, that any finite-dimensional C-contramodule
is a contramodule over a finite-dimensional subcoalgebra of C?
(The answer: this is not true. Let C be the coalgebra such that
the algebra C^* has the form C^* = ke_1 \oplus e_1V^*e_2 \oplus ke_2,
where e_1 and e_2 are idempotents, e_1 + e_2 = 1, and V is some
vector space. Then the category of C-comodules is equivalent to
the category of pairs of vector spaces (M_1,M_2) endowed with a map
M_2 \to V\ot M_1; while the category of C-contramodules is equivalent
to the category of pairs of vector spaces (P_1, P_2) endowed with
a map Hom(V,P_2) \to P_1. If V is infinite-dimensional, then there
are more finite-dimensional contramodules then finite-dimensional
comodules. Generally, if X and Y -- are two finite-dimensional
comodules over a coalgebra C, then
Ext^i_{C-contra}(X,Y) = (Ext^i_{C-comod}(X,Y))^**
is the double dual vector space. On the other hand, the classes of
irreducible C-contramodules and irreducible C-comodules are in
bijective correspondence -- see Addition at the end of the second
letter of the 2002 series. (Added in June 2006.))
End of digression.
There are two tensor product-type operations defined on comodules
and contramodules. Namely, if M is a right C-comodule and P is
a left C-contramodule, then the contratensor product M \ocn_C P
is defined as the quotient space of the vector space M\ot_k P by
the image of the map from M \ot_k Hom_k(C,P), where the latter map
is the difference of two: one comes from the contraaction on P,
the other one is equal to the composition
M \ot_k Hom_k(C,P) \to M\ot_k C \ot_k Hom_k(C,P) \to M\ot_k P,
where the first arrow is the coaction on M, while the second one is
the evaluation. If M is a D-C-bicomodule, then M\ocn_C P turns out
to be a left D-comodule. Furthermore, if M is a left C-comodule
and P is a left C-contramodule, then Cohom_C(M,P) is the quotient
space of Hom_k(M,P) by the image of the obvious map from
Hom_k(C\ot_k M, P) = Hom_k(M, Hom_k(C,P)). If M is
a C-D-bicomodule, then the space Cohom_C(M,P) acquires a left
D-contramodule structure.
It appears that no operation of (tensor) product of two
contramodules exists. So contramodules behave somewhat like
distributions (which can be multiplied by the usual functions
only, but cannot be multiplied with each other).
The most important identities involving the operations defined
above are the following ones. For the contratensor product:
(b) Hom_{D-comod}(M\ocn_C P, L)
= Hom_{C-contra}(P, Hom_{D-comod}(M,L))
for any D-C-bicomodule M, left D-comodule L, and left
C-contramodule P. It follows immediately from the formulas (a)
and (b) that the contratensor product with a free contramodule is
(c) M \ocn_C U^C = M \ot_k U.
The identity for the space of cohomomorphisms:
(d) Cohom_D(M\oc_C L, P) = Cohom_C(L, Cohom_D(M,P))
for any D-C-bicomodule M, left C-comodule L, and left
D-contramodule P. There are also formulas for the cohomomorphisms
from a cofree comodule or into a free contramodule:
(e) Cohom_C(C, P) = P;
(f) Cohom_C(M, U^C) = Hom_k(M,U).
In particular, one can see from (b) that there is a pair of adjoint
functors between the categories C-comod and C-contra :
L \mapsto Hom_{C-comod}(C,L) and P \mapsto C\ocn_C P.
It is clear from (c) that the restrictions of these two functors to
the full subcategories of cofree comodules and free contramodules
are mutually inverse equivalences between these subcategories.
Finally, the properties (d) and (e) mean that the Cohom operation
defines on the category (C-contra)^op a structure of right module
category over the tensor category C-bicomod-C (with the cotensor
product).
Now let S be an algebra in the category of bicomodules over
a coalgebra C. By a left S-contramodule one means an object of
the category opposite to the category of S-modules in
the above-mentioned right module category (C-contra)^op over
C-bicomod-C; in other words, a left contramodule M over
a semialgebra S is a left C-contramodule endowed with
a homomorphism of C-contramodules M \to Cohom_C(S,M) satisfying
the associativity and unitality conditions of the following form:
two maps from M to the space
Cohom_C(S\oc_C S, M) = Cohom_C(S, Cohom_C(S, M))
must coincide, while the composition
M \to Cohom_C(S,M) \to Cohom_C(C,M) = M,
where the second arrow is induced by the unit map of the algebra
S, must be the identity map.
Borderline cases: if S=C, then the S-contramodules are simply
the contramodules over the coalgebra C. If C=k, then S is
a usual algebra in Vect, and S-contramodules are the usual
S-modules (as explained above). Warning: if the coalgebra C is
finite-dimensional (or even S is finite-dimensional), then, even
though C-comodules and C-contramodules are the same things, but
S-modules (that is C-comodules with S-module structure -- those
that were discussed in the previous letter) and S-contramodules
-- are different things! One can say that left S-modules are
left A^#-modules, while left S-contramodules are left A-modules
(see below).
So, I have given the definition of the desired category. I think
that the arguments of the semi-infinite Ext functor should be:
- the first argument ("from") is left S-modules;
- the second argument ("to") is left S-contramodules.
Furthermore, if M is a right S-module and U is an arbitrary
vector space, then the vector space Hom_k(M,U) acquires a natural
structure of left S-contramodule. For such contramodules,
the following formula connecting the semi-infinite Ext and Tor
should hold: Ext^{\inf/2}(M_1, Hom_k(M_2,U))
= Hom_k(Tor_{\inf/2}(M_2, M_1), U).
Remark: the category of left modules over a semialgebra S is
abelian if S is cofree (or at least injective) over C on the right.
The category of left contramodules over S is abelian if S is cofree
(or injective) on the left.
In the remaining part of this text, three constructions are briefly
described: 1) the functor Cohom for S-modules and S-contramodules
-- the semi-infinite Ext should be its derived functor;
2) the equivalence between left A-modules and left S-contramodules
in the situation when the semialgebra S is produced from an algebra
A with a subalgebra N, i.e. S = C\ot_N A -- this is an analogue of
the assertions (iv-v) from the second letter and of the main result
of the third letter;
3) the equivalence between C-cofree left S-modules and C-free
left S-contramodules -- this is an analogue of the assertions
(vi-vii) from the second letter.
1) If M is a left S-module and P is a left S-contramodule, then
the vector space Cohom_S(M,P) is defined as the kernel of the map
Cohom_C(M,P) \to Cohom_C(S\oc M, P) = Cohom_C(M, Cohom_C(S,P))
equal to the difference of two maps, one arising from the S-module
structure on M, the other one from the S-contramodule structure on P.
2) Let A be a usual algebra and N its subalgebra such that A is
a free left N-module, let C be a coalgebra and f: N \to C^*
an algebra homomorphism with dense image, as in the third letter.
Assume that S = C\ot_N A is a right C-comodule; then S acquires
a structure of semialgebra over S. It was explained in the third
letter that the category of right modules over S is equivalent to
the category of right A-modules that are C-comodules. I claim
that there is an analogous equivalence for contramodules -- with
the difference that left A-modules are obtained in the result.
More precisely, as we know, there is a C^*-module structure on
any C-contramodule; hence there is the induced left N-module
structure. Furthermoe, for any left N-module M and any left
C-contramodule P there is a natural map
(g) Cohom_C(C\ot_N M, P) \to Hom_N(M,P),
which is an isomorphism if the module M is free or
the contramodule P is free.
Consider the following category: its objects are left A-modules P
endowed with an additional structure of a left C-contramodule.
These two structures must satisfy the following two compatibility
conditions: firstly, the structures of left N-module on P induced
from these two structures must coincide. Secondly, consider
the map P \to Hom_N(A,P) corresponding to the action map
A\ot_N P \to P. By construction, this is a homomorphism of
N-modules. It is required that it be a homomorphism of
C-contramodules, where the C-contramodule structure on the space
Hom_N(A,P) comes from its isomorphism (g) with the space
Cohom_C(S,P). The category so constructed is isomorphic to
the category of left contramodules over the semialgebra S by
the definition; one only needs to check that the two kinds of
associativity and unitality conditions agree with each other
under the isomorphism (g).
Notice that if the algebra N is finite-dimensional and C=N^*,
then the structures of left C-contramodule and left N-module are
equivalent, so the category of left contramodules over S turns
out to be isomorphic to the category of left A-modules -- as
promised in the introduction.
Exercise: suppose that we are in Serezha's situation, i.e.,
A and N are graded algebras, all the components of N are
finite-dimensional, and N_i=0 for i<<0. Let C=N^* be
the graded dual coalgebra, as in Example (2) above. Suppose
that one can construct a semialgebra S = C\ot_N A as described
above, that is, the N-bimodule A is "left admissible" in
the sense of the third letter. Assume additionally that
A = N\ot_k V as a left N-module, where the graded vector space V
is bounded above. Let M be a graded left A-module bounded below.
Then on the vector space \prod_i M_i there is a natural structure
of left contramodule over S extending the structure of a left
C-contramodule described in Example (2).
3) The equivalence of categories of cofree left C-comodules
and free left C-contramodules was constructed above:
M \mapsto Hom_{C-comod}(C,M) and P \mapsto C\ocn_C P.
It remains to check that, under this equivalence, structures
of S-module on a C-comodule correspond bijectively to structures
of S-contramodule on the corresponding C-contramodule.
The latter is true, at least, under the assumption that
the semialgebra S is cofree over C on the left and on the right.
The desired correspondence arises from the following isomorphism
of Hom spaces:
(h) Hom_{C-comod}(S\oc_C (C\ocn_C P), M) =
Hom_{C-contra}(P, Cohom_C(S,Hom_C(C,M)),
which holds for any left comodule M, any left contramodule P, and
any left and right cofree bicomodule S.
The formula (h), in turn, is a corollary of the formula (b) and
the following two properties of "associativity" of tensor operations.
Firstly, for any right C-comodule L, C-D-bicomodule E, and
left D-contramodule P there is a natural map
(j) (L\oc_C E) \ocn_D P \to L\oc_C (E \ocn_D P),
which is an isomorphism if L is cofree or P is free. Secondly,
for any left C-comodule L, D-C-bicomodule E, and left D-comodule M
there is a natural map
(k) Cohom_C(L, Hom_{D-comod}(E,M)) \to Hom_D(E\oc_C L, M),
which is an isomorphism if either L or M is cofree. To deduce (h),
one needs to substitute D=C, L=S and E=C.
OK, this is about all I have to tell.
Lenya.
P.S. Remark: By the way, here is another formula of similar kind.
For any D-C-bicomodule E, left C-contramodule P, and left
D-contramodule Q there is a natural map
(l) Cohom_D(E\ocn_C P, Q) \to Hom_{C-contra}(P, Cohom_D(E,Q)),
which is an isomorphism if P is free or Q is free.
Formulas (a-f) and (j-l) seem to exhaust the list of such
properties of operations depending on co- and contramodule
structures. To this list, one should add the formulas into which
structures of co/contramodules over a coalgebra and structures of
module over an algebra enter simultaneously. One such identity
was written down at the end of the third letter, formula (g) is
a particular case of another one, etc.
Addition (April 2006). Here is another useful operation on co-
and contramodules over a semialgebra: the contratensor product
over S. Let M be a right module over S and P a left contramodule
over S. By the definition, M\ocn_S P is the cokernel of the map
(M\oc_C S)\ocn_C P \to M\ocn_C P which is the difference of
the following two maps. The first one is the map induced by
the action M\oc_C S \to M. The second one is the composition
(M\oc_C S)\ocn_C P \to (M\oc_C S)\ocn_C Cohom_C(S,P) \to M\ocn_C P,
where the first map being composed comes from the contraaction of
S in P, while the second map being composed arises from the fact
that, as one can check, the composition M\oc_C S\ot_k Hom_k(S,P) \to
M\ot_k S\ot_k Hom_k(S,P) \to M\ot_k P \to M\ocn_C P factorizes
through the surjection M\oc_C S\ot_k Hom_k(S,P) \to
(M\oc_C S)\ocn_C Cohom_C(S,P). Notice that the contratensor
product over S is a right exact functor (at least, when S is
cofree over C on the left). Also it is not difficult to check that
(M\ocn_S P)^*=Hom_{S-contra}(P,M^*). Presumably, when C = N^*
(N finite-dimensional) and S = C\ot_N A, the operation \ocn_S
corresponds to the tensor product over A.
Russian original is here -- https://posic.livejournal.com/413.html
The continuation (two Summer 2002 letters) is here -- https://posic.livejournal.com/2774248.html
A series of letters about the semi-infinite (co)homology of
associative algebras
This is my April 2023 English translation of my Summer 2000 and
Summer 2002 letters to Roma Bezrukavnikov and Serezha Arkhipov
with Spring 2006 additions/comments.
This translation keeps most peculiarities of the terminology,
assumptions etc. of the original 2000-02 letters. The reader
will notice the usage of "free/cofree module" assumptions where
"projective/injective" would certainly suffice, etc. The notation
like A-mod stands for the category of all A-modules, not just
finitely generated or finite-dimensional ones.
One exception is that a particularly conflicting and outdated
early terminology of "comodule algebra" is replaced by my modern
term "semialgebra" (to avoid creating any further confusion).
The reader is also advised to keep in mind that what was called
"the derived category D-prime" in early 2000's is now known as
the coderived category, while what was then called "the derived
category D-second" is now known as the contraderived category.
Several misprints found in the Russian original have been
corrected in this translation.
From: Leonid Positselski < posic@mccme.ru >
To: roma, hippie
Date: Fri, 1 Sep 2000 09:40:45 +0400 (MSD)
Subject: demystifying A^#, first part (with corrections)
Lines: 65
Hi Roma and Serezha,
I have made some further corrections beyond those already mentioned,
all concerning the left/right issues, and also some additions.
Thanks for your attention etc.
I. Description of the algebra A^# in the case when A=N\otimes B,
where N and B are subalgebras. Uses the definition of A^# that
Serezha usually gives. Clarifies the infinite-dimensional
situation a little bit. Otherwise it is useless.
Let A be a graded algebra with two graded subalgebras N and B
such that N_i is finite-dimensional for all i, N_i=0 for i<<0,
and B_i=0 for i>>0. Assume that the multiplication map
N\otimes_k B \to A is an isomorphism. Then the algebra A can be
uniquely recovered from N, B and the "transposition" map
\phi: B\otimes N \to N\otimes B.
By "raising the indices" one can construct from the map \psi a map
\psi: N^*\otimes B \to Hom_k(N,B) = B\otimes N^*
(the latter equality follows from the conditions on the gradings
of N and B. Hom_k denotes the graded Hom space, i.e., the direct
sum of the spaces of homogeneous maps of various degrees).
Now I digress to recall Serezha's definition of the modules S
and S'. So S=N^*\otimes_N A and S'=Hom_{B-right}(A,B) (the right
B-module Hom). These are two right A-modules. There is a map
S \to S' between them,
n^*\otimes_N a \mapsto (a'\mapsto < n^*, aa' >)
where < n^*, - >: A \to B is the map pairing n^* with the first
tensor factor in A=N\otimes B.
We have S = N^*\otimes_k B = S' as vector spaces. I claim that
the A-module map S \to S' described above coincides with
the vector space map \psi described above. For Serezha's
definition of the algebra A^# to make sense, it is needed that
\psi be an isomorphism. I am assuming that.
Now consider the inverse map
\psi^{-1}: B\otimes N^* \to N^*\otimes B.
I want to "lower its indices" back. Then one obtains a certain map
\phi^#: N\otimes B \to Hom_k(N^*,B).
Unfortunately, now it does not follow from anything that the image
of \phi^# is contained in the subspace of maps N^*\to B of finite
rank, i.e., in B\otimes N \subset Hom_k(N^*,B). I claim that
Serezha's algebra A^# exists precisely when this is so, and in this
case the algebra A^# is contracted from the "transposition map"
\phi^#.
In the finite-dimensional case the latter condition is trivial, so
it is sufficient that the map \psi be invertible.
Now we can define the algebra A^# without using the modules S and S',
but using only the maps \phi and \phi^#. Specifically, for the
existence of an algebra with a given "transposition" map it is
necessary and sufficient that such map satisfies certain identities
(associativity). One can check that \phi and \phi^# satisfy these
identities simultaneously.
Lenya.
From: Leonid Positselski < posic@mccme.ru >
To: roma, hippie
Date: Fri, 1 Sep 2000 09:43:12 +0400 (MSD)
Subject: demystifying A^#, second part (with corrections)
Lines: 77
II. Description of the algebra A^# in the finite-dimensional case.
Uses neither the subalgebra B nor the grading. Only uses
the subalgebra N. I recommend it.
Let us start with a digression. Let N be a finite-dimensional
algebra (literally finite-dimensional; a graded locally
finite-dimensional would not do). Then N^* is a coalgebra, and
N-modules are the same things as N^*-comodules. On N-modules,
there is the operation of tensor product, while on N^*-comodules
there is the cotensor product. (If one wishes, the latter can be
defined in module terms as Hom_{N-bimod-N}(N,P\otimes_k Q),
where P and Q are a right and a left N-module, respectively.)
We will denote the tensor product of N-modules by P\ot_N Q and
the cotensor product by P\oc_{N^*} Q. On the category of
N-modules, there are (Serezha's) endofunctors P \mapsto N^*\ot_N P
and P \mapsto N\oc_{N^*}P = Hom_N(N^*,P).
I claim that for any right N-module P and left N-module Q there is
a map (of abelian groups)
(1) P\ot_N Q \to P\oc_{N^*} (N^*\ot_N Q),
which is an isomorphism, at least, when the module Q is free or
the module P is cofree. Analogously, there is a map
(2) P\ot_N (N\oc_{N^*} Q) \to P\oc_{N^*} Q,
and it is an isomorphism if the module Q is cofree or the module P
is free. (A kind of mutual associativity of the tensor and cotensor
products. A simple exercise.)
End of digression.
Now consider the tensor category of N-bimodules (with respect to
the tensor product). Bimodules that are free (say) on the left
form a tensor subcategory there. Analogously, there is the tensor
category of N-bimodules with respect to the cotensor product.
The left-cofree bimodules form a tensor subcategory in it. Then
the "associativity" formulas (1-2) mean that:
(i) the functor Q \mapsto N^*\ot_N Q is a tensor functor from
left-free bimodules w.r.t. the tensor product to left-cofree
bimodules w.r.t. the cotensor product;
(ii) Q \mapsto N\oc_{N^*} Q is a tensor functor in the opposite
direction;
(iii) these two functors are mutually inverse, and consequently,
equivalences of tensor categories.
In particular, these functors induce equivalences of the categories
of algebra objects in the first and the second tensor category.
Now let A be a (usual) algebra containing a subalgebra N; then A
is also an algebra in the category of N-bimodules. We see that if
A is free over N on the left, then S = N^*\ot_N A is a semialgebra
over N (i.e., an algebra with respect to the cotensor product).
Furthermore, if S turns out to be cofree on the right (and not only
on the left) over N, then one can define
A^# = S \oc_{N^*} N = (N^* \ot_N A) \oc_{N^*} N.
This is an algebra in the category of N-bimodules with respect to
the tensor product, i.e., simply an associative algebra containing N.
The condition that S is a cofree right N-module corresponds to
the condition of invertibility of the map \psi from my first letter.
The same formulas (1-2) imply that
(iv) right A-modules are the same things as right S-modules
(i.e. right N-modules with an S-module structure w.r.t. the cotensor
product);
(v) left A^#-modules are the same things as left S-modules;
(vi) the category of N-free left A-modules is equivalent to
the category of N-cofree left S-modules;
(vii) the category of N-free right A^#-modules is equivalent to
the category of N-cofree right S-modules.
Lenya.
From: Leonid Positselski < posic@mccme.ru >
To: roma, hippie
Date: Fri, 1 Sep 2000 09:59:50 +0400 (MSD)
Subject: demystifying A^#, third part (with corrections and additions)
Lines: 120
III. The semialgebra S and modules over it. A description using
neither the subalgebra B nor the grading. The infinite-dimensional
case.
As usual, I start with a digression. Let N be an algebra and C
a coalgebra (over the same field). I recall that there is a natural
algebra structure on the vector space C^* (such that any C-comodule
is a C^*-module). Suppose given an algebra homomorphism f: N\to C^*
with a dense image. In other words, there should be given a pairing
C\otimes N \to k that is compatible with the algebra and coalgebra
structures and nondegenerate in the second argument.
The datum of N, C and f is equivalent to the datum of a full
subcategory in the category of (say, left) N-modules with
the following properties: it is closed under subobjects and quotients
(but not necessarily under extensions!) and every module from this
subcategory is the union of its finite-dimensional submodules.
Specifically, to a triple (N, C, f) the fully faithful functor of
"restriction of scalars" f^*: C-comod \to N-mod is assigned.
A typical example: if H is an algebraic group (over a field of
characteristic 0), then there is the associated triple (N, C, f),
where N=U(h), h is the Lie algebra of H, and C=C(H) is the coalgebra
of functions on H. This triple corresponds to the full subcategory
in U(h)-mod consisting of the representations that can be integrated
to H.
Another example: if N is a graded algebra such that all N_i are
finite-dimensional and N_i=0 for i<<0, then one can take C to be
the graded dual vector space N^* to N. Then the N^*-comodules are
the N-modules with the property that for every element x from
such module one has N_i x = 0 for all i>>0.
A banal example: if N is finite-dimensional, one can take C=N^*.
Then all N-modules are C-comodules.
End of digression.
So, suppose given a triple (N, C, f). I will presume that
the categories of (left and right) C-comodules are embedded into
the respective categories of N-modules by means of f^*.
Accordingly, a statement like "such-and-such right N-module
is a C-comodule" means that this module belongs to the subcategory
of C-comodules in right N-modules.
Now in the tensor category of N-bimodules there is the following
tensor subcategory: a bimodule E belongs to this subcategory if for
any right C-comodule M the right N-module M\ot_N E is a C-comodule.
(It suffices that the N-module C\ot_N E be a right C-comodule.)
There is also the full subcategory, already familiar to us, of
N-bimodules that are free over N on the left. I am interested in
the intersection of these two full subcategories, i.e.,
the bimodules satisfying both the assumptions. I will call such
bimodules left (N,C,f)-admissible.
I claim that the functor E \mapsto C\ot_N E is a tensor functor
from left (N,C,f)-admissible bimodules to the bicomodules over C
(with respect to the cotensor product). So if N is a subalgebra
in A and A is left (N,C,f)-admissible, then S = C\ot_N A is
an algebra in the category of bicomodules over C. Moreover, let
M be a right C-comodule. Then to define a right A-module structure
on M (extending the right N-module structure related to the given
C-comodule structure) -- is the same thing as to define on M
a structure of right module, with respect to the cotensor product,
over the "semialgebra" S.
In other words, the full subcategory in the category of right
A-modules consisting of those modules which, viewed as N-modules,
are C-comodules, is isomorphic to the category of right modules
over S. This is some kind of infinite-dimensional generalization
of the property (iv) from the previous letter.
Both the claims are corollaries of the following "associativity"
for the tensor and cotensor products, generalizing formula (1): for
any right C-comodule P and left N-module Q there is a natural map
P \ot_N Q \to P\oc_C (C\ot_N Q),
which is an isomorphism if Q is free or P is cofree.
In fact, more generally, for any (unrelated) algebra N and
coalgebra C, comodule P and module Q, and a vector space T with
commuting structures of a left C-comodule and right N-module
there is a natural map
(P\oc_C T) \ot_N Q \to P \oc_C (T\ot_N Q),
which is an isomorphism if Q is free or P is cofree (while
no conditions are imposed on T).
Thus, as arguments of the functor of semi-infinite Tor one should
use:
- instead of bounded above graded right A-modules, while
the algebra N is positively graded, as Serezha had it --
right modules over the "semialgebra" S = C\ot_N A;
- instead of bounded above graded left A^#-modulej-- left
modules over S [see the previous letter, property (v)].
The final part will follow.
Lenya.
P.S. Remark: As in the previous letter, it is important here
that the semialgebra S should be C-cofree not only on the left
(which holds by construction), but also on the right. This is
needed, in particular, so that the category of left modules over S
were abelian (see also item 3 at the end of the fourth letter).
Suppose A = N\ot_k V as a left N-module, where V is some vector
space. Then S = C\ot_N A = C\ot_k V. The structure of right
C-comodule on S is given by a map S \to S\ot_k C. Denote by \psi_V
the composition of maps
C\ot_k V = S \to S\ot_k C = C\ot_k V\ot_k C \to V\ot_k C,
where the last arrow is obtained by applying the counit to the first
factor. As one can easily see, this is some kind of generalization
of the map \psi from the first letter. For S to be C-cofree on
on the right, it is obviously sufficient that there were a subspace
V \subset A for which \psi_V is invertible.
From: Leonid Positselski < posic@mccme.ru >
To: roma, hippie
Date: Fri, 1 Sep 2000 10:03:19 +0400 (MSD)
Subject: demystifying A^#, final part (with corrections and additions)
Lines: 364
IV. Contramodules. A description of the category where the second
argument of the semi-infinite Ext functor ranges. Nontrivial already
in the finite-dimensional case; works in the infinite-dimensional
case, too.
I start this final installment of these notes with a long
introduction (in an attempt to explain the problem to be solved).
In a reasonable sense, in the infinite-dimensional situation
the algebra A (or A^#) contains more information than
the semialgebra S. To construct A^# starting from S, as it was
explained in the first letter, one needs to have a subalgebra B.
If B is not given, one can only recover from S certain completions
of A and A^#: specifically,
A~ = End_{S-left}(S)^op and A^{#~} = End_{S-right}(S)
i.e., the endomorphisms of S as a left or right S-module. Indeed,
End_{S-right}(S) = End_{A-right}(S) = Hom_{N-right}(C, C\ot_N A)
= Hom_{C-right}(C, B\ot_k C) = Hom_k(C,B)
as a vector space (the first equality holds in view of
the equivalence of module categories from my previous letter,
the third one holds by virtue of a suitable version of
the isomorphism \psi).
In other words, it seems that from the semialgebra S one cannot
recover the whole category of right A-modules, but only
the "category O" -- the subcategory of those modules that are
C-comodules. The completion of the algebra A described above is
precisely the completion that acts on the modules from
"category O". (In the finite-dimensional case there are no
such distinctions, of course.)
I recall that (as Serezha taught us) the arguments of
the semi-infinite Ext functor are:
- The first argument ("from") -- left A^#-modules with
the grading bounded above. In the language of the semialgebra S,
these are left S-modules.
- The second argument ("to") -- left A-modules bounded below.
If the algebra N is infinite-dimensional, such modules are not even
C-comodules. So far now we did not know how to describe such
N-modules in terms of the coalgebra C, let alone describing
such A-modules in terms of the semialgebra S.
It will be explained below how to construct the abelian category
to which the second argument of the semi-infinite Ext functor
belongs in terms of the semialgebra S. This question is
surprizingly nontrivial in the finite-dimensional case already,
as -- following the assertions (iv-vii) from the second letter --
so far we could not recover the category of *left* A-modules (or
vice versa, right A^#-modules) in terms of S. We only know how to
recover the categories of right A-modules and left A^#-modules,
and also *N-free* left A-modules and right A^#-modules. Of course,
we want to use rather arbitrary modules and not just N-free ones
as arguments of the semi-infinite Ext. The abelian category of
"contramodules" which I will construct for a semialgebra S, will
coincide with the category of left A-modules in
the finite-dimensional case.
Surprizingly, one can even generalize the equivalences of
categories of N-free and N-cofree modules [statements (vi-vii) from
the second letter] to the infinite-dimensional case; so Roma's
definition of semi-infinite Ext can be likely generalized, too.
So, I start with a digression.
There are modules, there are comodules, and there are also
contramodules. For an infinite-dimensional coalgebra C one can
construct, alongside with the categories of left and right
comodules, two other abelian categories -- of left and right
contramodules. This is done in the following way.
Vector spaces (over a fixed field k) form a tensor catgory.
Therefore, one can consider algebras or coalgebras in this tensor
category; these are the usual algebras and coalgebras. Now if
one is interested in modules or comodules over such algebras or
coalgebras, one can define such (co)modules as objects of the same
category Vect. These will be the usual modules or comodules.
But one can also choose some *module category* over the tensor
category Vect, that is, a category M together with an "action"
functor Vect\times M \to M, with the suitable associativity
constraint -- and consider objects of this module category endowed
with a (co)module structure over a fixed (co)algebra from Vect.
On the category Vect^op opposite to Vect there is a structure of
(left) module category over Vect given by the formula
(V, W^op) \mapsto Hom_k(V,W)^op.
This is a module category in view of the isomorphism
Hom(U\ot_k V, W) = Hom(U, Hom(V,W)).
To obtain a right module category structure, one needs to use
another isomorphism
Hom(U\ot_k V, W) = Hom(V, Hom(U,W)).
in the role of the associativity constraint.
Now if B is an algebra in Vect, then a B-module in the module
category Vect^op over Vect is a called a contramodule over B.
More precisely, by a left B-contramodule one means an object of
the category opposite to the category of modules over B in
the *right* module category Vect^op (so that the forgetful functor
from contramodules to Vect is covariant). Simply put, a contramodule
is a vector space V together with a map V \to Hom_k(B,V), which must
satisfy certain associativity and unitality conditions. However,
to specify a map V \to Hom_k(B,V) is the same a to specify a map
B\ot_k V \to V; hence left B-contramodules are the same things as
left B-modules.
For coalgebras, the situation is different. The definition is
the same: if C is a coalgebra in Vect, then a left contramodule
over C is an object of the category opposite to the category of
comodules over C in the right module category Vect^op over Vect.
In other words, a contramodule over C is a vector space V together
with a map
Hom_k(C,V) \to V,
satisfying conditions of coassociativity and counitality of
the following form: two maps from the vector space
Hom_k(C, Hom_k(C,V)) = Hom_k(C\ot C, V)
to the space V -- one obtained by iterating the "contraaction"
operation and the other one depending on the comultiplication on C
-- must coincide; the composition of maps V \to Hom_k(C,V) \to V,
where the first arrow comes from the counit on C and the second one
is the "contraaction", must be the identity map.
On any C-contramodule (just as on any C-comodule) there is
a natural structure of a module over the algebra C^*, since
the space C^*\ot_k V is embedded into Hom_k(C,V). All the three
categories (left C-comodules, left C-contramodules, and left
C^*-modules) agree if the coalgebra C is finite-dimensional;
otherwise, they are all different. (In particular, even though
the operations of infinite direct sum and infinite product exist
on all the three categories, the functor from C-comodules to
C^*-modules only preserves the infinite direct sums, while
the functor from C-contramodules to C^*-modules only preserves
the infinite products, but not vice versa.)
Examples: (1) If M is a right comodule over C and U is an arbitrary
vector space, then the space Hom_k(M,U) has a natural structure of
left contramodule over C. More generally, if there is a structure
of left D-(co)module on M commuting with the structure of right
C-comodule, and U is some D-(co)module, then Hom_D(M,U) is a left
contramodule over C.
(2) Let N be a graded algebra with finite-dimensional components
such that N_i=0 dlya i<<0, and let N^* be the graded dual coalgebra.
Then if M is a graded left N-module bounded below, then on
the direct product of the grading components of the module M there
is a natural structure of left contramodule over N^*.
Contramodules of the form Hom_k(C,U), where the coalgebra C is
viewed as a right comodule over itself and U is an arbitrary vector
space, are called free C-contramodules. A short notation:
Hom_k(C,U) =: U^C. Exercise: there is a natural isomorphism
(a) Hom_{C-contra}(U^C, P) = Hom_k(U,P)
for any left C-contramodule P.
The next exercise I cannot even solve myself: is it true that
finite-dimensional contramodules over an infinite-dimenional
coalgebra C are the same things as finite-dimensional comodules;
that is, in other words, that any finite-dimensional C-contramodule
is a contramodule over a finite-dimensional subcoalgebra of C?
(The answer: this is not true. Let C be the coalgebra such that
the algebra C^* has the form C^* = ke_1 \oplus e_1V^*e_2 \oplus ke_2,
where e_1 and e_2 are idempotents, e_1 + e_2 = 1, and V is some
vector space. Then the category of C-comodules is equivalent to
the category of pairs of vector spaces (M_1,M_2) endowed with a map
M_2 \to V\ot M_1; while the category of C-contramodules is equivalent
to the category of pairs of vector spaces (P_1, P_2) endowed with
a map Hom(V,P_2) \to P_1. If V is infinite-dimensional, then there
are more finite-dimensional contramodules then finite-dimensional
comodules. Generally, if X and Y -- are two finite-dimensional
comodules over a coalgebra C, then
Ext^i_{C-contra}(X,Y) = (Ext^i_{C-comod}(X,Y))^**
is the double dual vector space. On the other hand, the classes of
irreducible C-contramodules and irreducible C-comodules are in
bijective correspondence -- see Addition at the end of the second
letter of the 2002 series. (Added in June 2006.))
End of digression.
There are two tensor product-type operations defined on comodules
and contramodules. Namely, if M is a right C-comodule and P is
a left C-contramodule, then the contratensor product M \ocn_C P
is defined as the quotient space of the vector space M\ot_k P by
the image of the map from M \ot_k Hom_k(C,P), where the latter map
is the difference of two: one comes from the contraaction on P,
the other one is equal to the composition
M \ot_k Hom_k(C,P) \to M\ot_k C \ot_k Hom_k(C,P) \to M\ot_k P,
where the first arrow is the coaction on M, while the second one is
the evaluation. If M is a D-C-bicomodule, then M\ocn_C P turns out
to be a left D-comodule. Furthermore, if M is a left C-comodule
and P is a left C-contramodule, then Cohom_C(M,P) is the quotient
space of Hom_k(M,P) by the image of the obvious map from
Hom_k(C\ot_k M, P) = Hom_k(M, Hom_k(C,P)). If M is
a C-D-bicomodule, then the space Cohom_C(M,P) acquires a left
D-contramodule structure.
It appears that no operation of (tensor) product of two
contramodules exists. So contramodules behave somewhat like
distributions (which can be multiplied by the usual functions
only, but cannot be multiplied with each other).
The most important identities involving the operations defined
above are the following ones. For the contratensor product:
(b) Hom_{D-comod}(M\ocn_C P, L)
= Hom_{C-contra}(P, Hom_{D-comod}(M,L))
for any D-C-bicomodule M, left D-comodule L, and left
C-contramodule P. It follows immediately from the formulas (a)
and (b) that the contratensor product with a free contramodule is
(c) M \ocn_C U^C = M \ot_k U.
The identity for the space of cohomomorphisms:
(d) Cohom_D(M\oc_C L, P) = Cohom_C(L, Cohom_D(M,P))
for any D-C-bicomodule M, left C-comodule L, and left
D-contramodule P. There are also formulas for the cohomomorphisms
from a cofree comodule or into a free contramodule:
(e) Cohom_C(C, P) = P;
(f) Cohom_C(M, U^C) = Hom_k(M,U).
In particular, one can see from (b) that there is a pair of adjoint
functors between the categories C-comod and C-contra :
L \mapsto Hom_{C-comod}(C,L) and P \mapsto C\ocn_C P.
It is clear from (c) that the restrictions of these two functors to
the full subcategories of cofree comodules and free contramodules
are mutually inverse equivalences between these subcategories.
Finally, the properties (d) and (e) mean that the Cohom operation
defines on the category (C-contra)^op a structure of right module
category over the tensor category C-bicomod-C (with the cotensor
product).
Now let S be an algebra in the category of bicomodules over
a coalgebra C. By a left S-contramodule one means an object of
the category opposite to the category of S-modules in
the above-mentioned right module category (C-contra)^op over
C-bicomod-C; in other words, a left contramodule M over
a semialgebra S is a left C-contramodule endowed with
a homomorphism of C-contramodules M \to Cohom_C(S,M) satisfying
the associativity and unitality conditions of the following form:
two maps from M to the space
Cohom_C(S\oc_C S, M) = Cohom_C(S, Cohom_C(S, M))
must coincide, while the composition
M \to Cohom_C(S,M) \to Cohom_C(C,M) = M,
where the second arrow is induced by the unit map of the algebra
S, must be the identity map.
Borderline cases: if S=C, then the S-contramodules are simply
the contramodules over the coalgebra C. If C=k, then S is
a usual algebra in Vect, and S-contramodules are the usual
S-modules (as explained above). Warning: if the coalgebra C is
finite-dimensional (or even S is finite-dimensional), then, even
though C-comodules and C-contramodules are the same things, but
S-modules (that is C-comodules with S-module structure -- those
that were discussed in the previous letter) and S-contramodules
-- are different things! One can say that left S-modules are
left A^#-modules, while left S-contramodules are left A-modules
(see below).
So, I have given the definition of the desired category. I think
that the arguments of the semi-infinite Ext functor should be:
- the first argument ("from") is left S-modules;
- the second argument ("to") is left S-contramodules.
Furthermore, if M is a right S-module and U is an arbitrary
vector space, then the vector space Hom_k(M,U) acquires a natural
structure of left S-contramodule. For such contramodules,
the following formula connecting the semi-infinite Ext and Tor
should hold: Ext^{\inf/2}(M_1, Hom_k(M_2,U))
= Hom_k(Tor_{\inf/2}(M_2, M_1), U).
Remark: the category of left modules over a semialgebra S is
abelian if S is cofree (or at least injective) over C on the right.
The category of left contramodules over S is abelian if S is cofree
(or injective) on the left.
In the remaining part of this text, three constructions are briefly
described: 1) the functor Cohom for S-modules and S-contramodules
-- the semi-infinite Ext should be its derived functor;
2) the equivalence between left A-modules and left S-contramodules
in the situation when the semialgebra S is produced from an algebra
A with a subalgebra N, i.e. S = C\ot_N A -- this is an analogue of
the assertions (iv-v) from the second letter and of the main result
of the third letter;
3) the equivalence between C-cofree left S-modules and C-free
left S-contramodules -- this is an analogue of the assertions
(vi-vii) from the second letter.
1) If M is a left S-module and P is a left S-contramodule, then
the vector space Cohom_S(M,P) is defined as the kernel of the map
Cohom_C(M,P) \to Cohom_C(S\oc M, P) = Cohom_C(M, Cohom_C(S,P))
equal to the difference of two maps, one arising from the S-module
structure on M, the other one from the S-contramodule structure on P.
2) Let A be a usual algebra and N its subalgebra such that A is
a free left N-module, let C be a coalgebra and f: N \to C^*
an algebra homomorphism with dense image, as in the third letter.
Assume that S = C\ot_N A is a right C-comodule; then S acquires
a structure of semialgebra over S. It was explained in the third
letter that the category of right modules over S is equivalent to
the category of right A-modules that are C-comodules. I claim
that there is an analogous equivalence for contramodules -- with
the difference that left A-modules are obtained in the result.
More precisely, as we know, there is a C^*-module structure on
any C-contramodule; hence there is the induced left N-module
structure. Furthermoe, for any left N-module M and any left
C-contramodule P there is a natural map
(g) Cohom_C(C\ot_N M, P) \to Hom_N(M,P),
which is an isomorphism if the module M is free or
the contramodule P is free.
Consider the following category: its objects are left A-modules P
endowed with an additional structure of a left C-contramodule.
These two structures must satisfy the following two compatibility
conditions: firstly, the structures of left N-module on P induced
from these two structures must coincide. Secondly, consider
the map P \to Hom_N(A,P) corresponding to the action map
A\ot_N P \to P. By construction, this is a homomorphism of
N-modules. It is required that it be a homomorphism of
C-contramodules, where the C-contramodule structure on the space
Hom_N(A,P) comes from its isomorphism (g) with the space
Cohom_C(S,P). The category so constructed is isomorphic to
the category of left contramodules over the semialgebra S by
the definition; one only needs to check that the two kinds of
associativity and unitality conditions agree with each other
under the isomorphism (g).
Notice that if the algebra N is finite-dimensional and C=N^*,
then the structures of left C-contramodule and left N-module are
equivalent, so the category of left contramodules over S turns
out to be isomorphic to the category of left A-modules -- as
promised in the introduction.
Exercise: suppose that we are in Serezha's situation, i.e.,
A and N are graded algebras, all the components of N are
finite-dimensional, and N_i=0 for i<<0. Let C=N^* be
the graded dual coalgebra, as in Example (2) above. Suppose
that one can construct a semialgebra S = C\ot_N A as described
above, that is, the N-bimodule A is "left admissible" in
the sense of the third letter. Assume additionally that
A = N\ot_k V as a left N-module, where the graded vector space V
is bounded above. Let M be a graded left A-module bounded below.
Then on the vector space \prod_i M_i there is a natural structure
of left contramodule over S extending the structure of a left
C-contramodule described in Example (2).
3) The equivalence of categories of cofree left C-comodules
and free left C-contramodules was constructed above:
M \mapsto Hom_{C-comod}(C,M) and P \mapsto C\ocn_C P.
It remains to check that, under this equivalence, structures
of S-module on a C-comodule correspond bijectively to structures
of S-contramodule on the corresponding C-contramodule.
The latter is true, at least, under the assumption that
the semialgebra S is cofree over C on the left and on the right.
The desired correspondence arises from the following isomorphism
of Hom spaces:
(h) Hom_{C-comod}(S\oc_C (C\ocn_C P), M) =
Hom_{C-contra}(P, Cohom_C(S,Hom_C(C,M)),
which holds for any left comodule M, any left contramodule P, and
any left and right cofree bicomodule S.
The formula (h), in turn, is a corollary of the formula (b) and
the following two properties of "associativity" of tensor operations.
Firstly, for any right C-comodule L, C-D-bicomodule E, and
left D-contramodule P there is a natural map
(j) (L\oc_C E) \ocn_D P \to L\oc_C (E \ocn_D P),
which is an isomorphism if L is cofree or P is free. Secondly,
for any left C-comodule L, D-C-bicomodule E, and left D-comodule M
there is a natural map
(k) Cohom_C(L, Hom_{D-comod}(E,M)) \to Hom_D(E\oc_C L, M),
which is an isomorphism if either L or M is cofree. To deduce (h),
one needs to substitute D=C, L=S and E=C.
OK, this is about all I have to tell.
Lenya.
P.S. Remark: By the way, here is another formula of similar kind.
For any D-C-bicomodule E, left C-contramodule P, and left
D-contramodule Q there is a natural map
(l) Cohom_D(E\ocn_C P, Q) \to Hom_{C-contra}(P, Cohom_D(E,Q)),
which is an isomorphism if P is free or Q is free.
Formulas (a-f) and (j-l) seem to exhaust the list of such
properties of operations depending on co- and contramodule
structures. To this list, one should add the formulas into which
structures of co/contramodules over a coalgebra and structures of
module over an algebra enter simultaneously. One such identity
was written down at the end of the third letter, formula (g) is
a particular case of another one, etc.
Addition (April 2006). Here is another useful operation on co-
and contramodules over a semialgebra: the contratensor product
over S. Let M be a right module over S and P a left contramodule
over S. By the definition, M\ocn_S P is the cokernel of the map
(M\oc_C S)\ocn_C P \to M\ocn_C P which is the difference of
the following two maps. The first one is the map induced by
the action M\oc_C S \to M. The second one is the composition
(M\oc_C S)\ocn_C P \to (M\oc_C S)\ocn_C Cohom_C(S,P) \to M\ocn_C P,
where the first map being composed comes from the contraaction of
S in P, while the second map being composed arises from the fact
that, as one can check, the composition M\oc_C S\ot_k Hom_k(S,P) \to
M\ot_k S\ot_k Hom_k(S,P) \to M\ot_k P \to M\ocn_C P factorizes
through the surjection M\oc_C S\ot_k Hom_k(S,P) \to
(M\oc_C S)\ocn_C Cohom_C(S,P). Notice that the contratensor
product over S is a right exact functor (at least, when S is
cofree over C on the left). Also it is not difficult to check that
(M\ocn_S P)^*=Hom_{S-contra}(P,M^*). Presumably, when C = N^*
(N finite-dimensional) and S = C\ot_N A, the operation \ocn_S
corresponds to the tensor product over A.