Как перевести нашу старую работу с Сашей В. с языка когомологий на язык соотношений
Ср. http://posic.livejournal.com/655090.html
Dear Peter,
Thank you for your letter. I am glad that our old paper with
Vishik is now receiving some attention in the braid group circles.
Our paper is written in the cohomological language, but this
can be translated into the language of generators and relations.
In fact, our Main Theorem is a close analogue of the nonhomogeneous
Poincare-Birkhoff-Witt theorem of the kind concidered in Chapter 5
of the book "Quadratic algebras", the difference being that in
Chapter 5 we do algebras with nonhomogeneous relations with
components of the degrees 2, 1, 0, while the Main Theorem in our
paper with Vishik concerns algebras with nonhomogeneous relations
with components of the degrees 2, 3, 4, ... up to infinity.
Let me elaborate on the translation procedure. The basic facts are
that, for nilpotent-type structures (positively graded, augmented
and complete with respect to the augmentation filtration, etc.),
the first cohomology with constant coefficients describe generators,
the second cohomology describe relations, and the third cohomology
describe (first) syzygies between relations. More precisely,
the cohomology spaces are the dual vector spaces to the spaces of
generators, relations, or syzygies (the cohomology depend covariantly
on the coalgebra C, but it is itself dual to the algebra K whose
generators and relations we are discussing).
In particular, the condition (1) from our Main Theorem, claiming
that H^2(C) is multiplicatively generated by H^1(C), is equivalent
(mutatis mutandis) to what you call "the assumptions of
Proposition 6" in your paper, namely, that the relations are
linearly independent modulo the cube of the augmentation ideal
of the free algebra. Basically, on the space of relations you
have the comultiplication map H_2 \to H_1\otimes H_1, H_1 being
the space of generators. The kernel of this map corresponds
precisely to those relations (or linear combinations of
the relations) that lie in the cube of the augmentation ideal.
The comultiplication map itself assigns to any relation its quadratic
part, considered as a quadratic expression in terms of the generators.
To translate the somewhat more complicated condition (2) from our
theorem (claiming that there are no cubic relations in cohomology)
to the language of generators and relations, one has to write down
the spectral sequence used in the proof in our paper. As we explain,
in our paper, the cohomology of the graded (co)algebra grC is
a bigraded algebra H = \bigoplus H^{ij}(grC), with the cohomological
grading i and the internal grading j. One has i\le j for all
nonzero components, and the diagonal subalgebra \bigoplus H^{ii}
is particularly nice, being always quadratic and quadratically dual
to the quadratic algebra q(grC). The space H^{2,2}(grC) is dual to
the space of quadratic relations in grC (or more precisely in grK),
which is the same thing as the space of relations in q(grC); and
the spaces H^{2,j}(grC) for j>2 are the dual spaces to the spaces of
higher-degree relations in grC (or grK).
Now let us look on the spectral sequence H^{ij}(grC) => H^i(C).
Since grC is by definition generated in degree 1, we always have
H^{1,j}(grC) = 0 for j\ne 1. Our condition (1) (i.e. what you call
the assumptions of Proposition 6) basically means that all
the terms H^{2,j}(grC) have to die in the spectral sequence
(as H^2(C) is generated by H^1(C) and H^1(C) lies in component N_1
of the increasing filtration N on H^*(C) related to this spectral
sequence, so H^2(C) has to lie in N_2). The only way the elements
of the terms H^{2,j}(grC) can die in this this spectral sequence is
by being taken by some differentials to nonzero elements in
the terms H^{3,s}(grC). The differentials go d_r: H^{2,j}(grC) \to
H^{3,j-r}(grC), where r\ge 1.
This is how the syzygies of q(grC) come into play here. The spaces
H^{3,s}(grC) are dual to the syzygies in grC rather than q(grC),
but there is some induction procedure to relate one to the other.
Basically, given a graded (co)algebra D, the space H^{3,n}(D) only
depends on the components D_1, ..., D_{n-1}. Hence assuming there
are no relations in grC of degrees 3, ..., n-1, one can be sure
that H^{3,n}(grC) = H^{3,n}(q(grC)).
Furthermore, what you call "a precise way in which syzygies of K
induce syzygies of q(grK)" in your letter, I call the natural
filtration on H^3(C) with the successive quotients being certain
subquotients of H^{3,n}(grC) (as usually in a spectral sequence).
When you say that "all syzygies of q(grK) also hold in K", I say
that all the differentials coming into H^{3,s}(grC) in my spectral
sequence vanish, so one does not have to pass to quotient spaces of
H^{3,s}(grC) in order to obtain (the successive quotients to
a filtraion on) H^3(C). Let me repeat that the 3rd cohomology
spaces are dual to the syzygy spaces, so passing to quotient spaces
of the 3rd cohomology would mean passing to subspaces of
the syzygies of grC.
Let us consider the situation in the internal grading 3 (which is
singularly important in the Koszul case). Since the term H^{3,3}(grC)
corresponds to the lowest term of the increasing filtraion on H^3(C)
and all the differentials going out of H^{3,3}(grC) obviously vanish,
there is a natural map H^{3,3}(grC) \to H^3(C). This is the dual map
to your "way in which syzygies of K induce syzygies of q(grC)" (in
degree 3). The condition (2) from our Main Theorem says precisely
that this map is injective. In the dual picture this would mean that
the natural map from the syzygies of C to (the third graded piece of)
the syzygies of grC must be surjective. (Let me also reiterate that
H^{3,3}(grC) is always the same as H^{3,3}(q(grC)).)
Indeed, it is explained in the proof in our paper that the diagonal
cohomology \bigoplus H^{i,i}(grC) always coincide with the quadratic
part of (or what you would call the quadratic approximation to)
the cohomology of C. So injectivity of the map H^{3,3}(grC) \to
H^3(C) precisely means that there are no nontrivial cubic relations
in H^*(C). From the point of view of the spectral sequence, this
means that all the differentials coming into H^{3,3}(grC) vanish.
In the Koszul case, the induction procedure (as explained above)
proves that vanishing of these differentials implies vanishing of
all the other differentials coming into H^3(grC) (and indeed all
the other differentials in the spectral sequence), assuming that
the condition (1) also holds.
To sum up: the condition (1) in our Main Theorem (concerning generation
of H^2) is equivalent to your condition that any linear combination of
the relations in K has a nonzero quadratic part. Our condition (2)
(concerning there being no cubic relations in H^3) is equivalent to
your condition that all the syzygies of degree 3 in q(grK) also hold
in K. Finally, our Koszulity condition (3) on the quadratic part of
the cohomology algebra H(C) is equivalent to your Koszulity condition
on the quadratic part of the associated graded algebra q(grK), as
these two quadratic algebras are always quadratic dual to each other.
As you mention, in these assumptions we prove, among other things, that
the coalgebra grC is quadratic and Koszul. So (after all the above
translations from one language to the other one are done) we indeed
prove what you've tentatively said we prove.
There is one caveat. When I say "equivalent" above, it always means
"equivalent in the finitely generated case, or up to replacing
discrete vector spaces with compact vector spaces". The point is,
the topological dual vector space to the augmentation completion of
an augmented algebra is a (co)nilpotent coalgebra whenever the algebra
was finitely generated. Conversely, the dual vector space to
a conilpotent coalgebra is a quotient algebra of the augmentation
completion of a free associative algebra whenever the coalgebra was
finitely cogenerated.
In the infinitely generated case, these are just two different
categories, one of them (yours) consisting of algebras with discrete
infinite-dimensional spaces of generators, the other one (ours)
consisting of algebras with compact (i.e., topological dual to
discrete) spaces of generators. I would hesitate to claim without
further thinking that our theorem and/or its proof applies to
quotient algebras of augmentation completions of infinitely generated
discrete free associative algebras.
With best wishes,
Leonid.
Dear Peter,
Thank you for your letter. I am glad that our old paper with
Vishik is now receiving some attention in the braid group circles.
Our paper is written in the cohomological language, but this
can be translated into the language of generators and relations.
In fact, our Main Theorem is a close analogue of the nonhomogeneous
Poincare-Birkhoff-Witt theorem of the kind concidered in Chapter 5
of the book "Quadratic algebras", the difference being that in
Chapter 5 we do algebras with nonhomogeneous relations with
components of the degrees 2, 1, 0, while the Main Theorem in our
paper with Vishik concerns algebras with nonhomogeneous relations
with components of the degrees 2, 3, 4, ... up to infinity.
Let me elaborate on the translation procedure. The basic facts are
that, for nilpotent-type structures (positively graded, augmented
and complete with respect to the augmentation filtration, etc.),
the first cohomology with constant coefficients describe generators,
the second cohomology describe relations, and the third cohomology
describe (first) syzygies between relations. More precisely,
the cohomology spaces are the dual vector spaces to the spaces of
generators, relations, or syzygies (the cohomology depend covariantly
on the coalgebra C, but it is itself dual to the algebra K whose
generators and relations we are discussing).
In particular, the condition (1) from our Main Theorem, claiming
that H^2(C) is multiplicatively generated by H^1(C), is equivalent
(mutatis mutandis) to what you call "the assumptions of
Proposition 6" in your paper, namely, that the relations are
linearly independent modulo the cube of the augmentation ideal
of the free algebra. Basically, on the space of relations you
have the comultiplication map H_2 \to H_1\otimes H_1, H_1 being
the space of generators. The kernel of this map corresponds
precisely to those relations (or linear combinations of
the relations) that lie in the cube of the augmentation ideal.
The comultiplication map itself assigns to any relation its quadratic
part, considered as a quadratic expression in terms of the generators.
To translate the somewhat more complicated condition (2) from our
theorem (claiming that there are no cubic relations in cohomology)
to the language of generators and relations, one has to write down
the spectral sequence used in the proof in our paper. As we explain,
in our paper, the cohomology of the graded (co)algebra grC is
a bigraded algebra H = \bigoplus H^{ij}(grC), with the cohomological
grading i and the internal grading j. One has i\le j for all
nonzero components, and the diagonal subalgebra \bigoplus H^{ii}
is particularly nice, being always quadratic and quadratically dual
to the quadratic algebra q(grC). The space H^{2,2}(grC) is dual to
the space of quadratic relations in grC (or more precisely in grK),
which is the same thing as the space of relations in q(grC); and
the spaces H^{2,j}(grC) for j>2 are the dual spaces to the spaces of
higher-degree relations in grC (or grK).
Now let us look on the spectral sequence H^{ij}(grC) => H^i(C).
Since grC is by definition generated in degree 1, we always have
H^{1,j}(grC) = 0 for j\ne 1. Our condition (1) (i.e. what you call
the assumptions of Proposition 6) basically means that all
the terms H^{2,j}(grC) have to die in the spectral sequence
(as H^2(C) is generated by H^1(C) and H^1(C) lies in component N_1
of the increasing filtration N on H^*(C) related to this spectral
sequence, so H^2(C) has to lie in N_2). The only way the elements
of the terms H^{2,j}(grC) can die in this this spectral sequence is
by being taken by some differentials to nonzero elements in
the terms H^{3,s}(grC). The differentials go d_r: H^{2,j}(grC) \to
H^{3,j-r}(grC), where r\ge 1.
This is how the syzygies of q(grC) come into play here. The spaces
H^{3,s}(grC) are dual to the syzygies in grC rather than q(grC),
but there is some induction procedure to relate one to the other.
Basically, given a graded (co)algebra D, the space H^{3,n}(D) only
depends on the components D_1, ..., D_{n-1}. Hence assuming there
are no relations in grC of degrees 3, ..., n-1, one can be sure
that H^{3,n}(grC) = H^{3,n}(q(grC)).
Furthermore, what you call "a precise way in which syzygies of K
induce syzygies of q(grK)" in your letter, I call the natural
filtration on H^3(C) with the successive quotients being certain
subquotients of H^{3,n}(grC) (as usually in a spectral sequence).
When you say that "all syzygies of q(grK) also hold in K", I say
that all the differentials coming into H^{3,s}(grC) in my spectral
sequence vanish, so one does not have to pass to quotient spaces of
H^{3,s}(grC) in order to obtain (the successive quotients to
a filtraion on) H^3(C). Let me repeat that the 3rd cohomology
spaces are dual to the syzygy spaces, so passing to quotient spaces
of the 3rd cohomology would mean passing to subspaces of
the syzygies of grC.
Let us consider the situation in the internal grading 3 (which is
singularly important in the Koszul case). Since the term H^{3,3}(grC)
corresponds to the lowest term of the increasing filtraion on H^3(C)
and all the differentials going out of H^{3,3}(grC) obviously vanish,
there is a natural map H^{3,3}(grC) \to H^3(C). This is the dual map
to your "way in which syzygies of K induce syzygies of q(grC)" (in
degree 3). The condition (2) from our Main Theorem says precisely
that this map is injective. In the dual picture this would mean that
the natural map from the syzygies of C to (the third graded piece of)
the syzygies of grC must be surjective. (Let me also reiterate that
H^{3,3}(grC) is always the same as H^{3,3}(q(grC)).)
Indeed, it is explained in the proof in our paper that the diagonal
cohomology \bigoplus H^{i,i}(grC) always coincide with the quadratic
part of (or what you would call the quadratic approximation to)
the cohomology of C. So injectivity of the map H^{3,3}(grC) \to
H^3(C) precisely means that there are no nontrivial cubic relations
in H^*(C). From the point of view of the spectral sequence, this
means that all the differentials coming into H^{3,3}(grC) vanish.
In the Koszul case, the induction procedure (as explained above)
proves that vanishing of these differentials implies vanishing of
all the other differentials coming into H^3(grC) (and indeed all
the other differentials in the spectral sequence), assuming that
the condition (1) also holds.
To sum up: the condition (1) in our Main Theorem (concerning generation
of H^2) is equivalent to your condition that any linear combination of
the relations in K has a nonzero quadratic part. Our condition (2)
(concerning there being no cubic relations in H^3) is equivalent to
your condition that all the syzygies of degree 3 in q(grK) also hold
in K. Finally, our Koszulity condition (3) on the quadratic part of
the cohomology algebra H(C) is equivalent to your Koszulity condition
on the quadratic part of the associated graded algebra q(grK), as
these two quadratic algebras are always quadratic dual to each other.
As you mention, in these assumptions we prove, among other things, that
the coalgebra grC is quadratic and Koszul. So (after all the above
translations from one language to the other one are done) we indeed
prove what you've tentatively said we prove.
There is one caveat. When I say "equivalent" above, it always means
"equivalent in the finitely generated case, or up to replacing
discrete vector spaces with compact vector spaces". The point is,
the topological dual vector space to the augmentation completion of
an augmented algebra is a (co)nilpotent coalgebra whenever the algebra
was finitely generated. Conversely, the dual vector space to
a conilpotent coalgebra is a quotient algebra of the augmentation
completion of a free associative algebra whenever the coalgebra was
finitely cogenerated.
In the infinitely generated case, these are just two different
categories, one of them (yours) consisting of algebras with discrete
infinite-dimensional spaces of generators, the other one (ours)
consisting of algebras with compact (i.e., topological dual to
discrete) spaces of generators. I would hesitate to claim without
further thinking that our theorem and/or its proof applies to
quotient algebras of augmentation completions of infinitely generated
discrete free associative algebras.
With best wishes,
Leonid.