Mixing
![How to get more time to wait for other job offer ? [duplicate]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgjbpfN9tAutmK93bJRC3ZoROZzi2TJDms5n8_qJuhgE0a9b52OOHayv3NGT8igAdFL7byXNst-_1DZK5SjrIJ28_6RQPUpBROqMs5s6jo-ZsjX8kjDwfxJufIitH3TaQRXWaGSQKRQib-f/s72-c/1.jpg)
Hochschild homology with coefficients in a certain bimodule
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
Let $A$ be a finite-dimensional $k$-algebra and $U$ and $V$ two finite-dimensional projective $A$-modules (maybe neither the finiteness nor projectivity has to play a role, but these requirements are satisfied in my problem).
Now consider the $A$-bimodule
beginalign
M := mathrmHom_A (U,A) otimes mathrmHom_A (A,V) ,
endalign where $A$ is an $A$-module via left multiplication making $mathrmHom_A (U,A)$ an $A$-module and $mathrmHom_A (A,V)$ an $A^textopp$-module.
I would like to compute the Hochschild homology $HH_*(A;M)$.
If you write down the corresponding Hochschild complex, there is an augmentation by $mathrmHom_A (U,V)$. Does that induce a quasi-isomorphism?
Maybe such a result is known to experts?
Is there, more generally, a coefficient theorem I could use?
As usual, thank you for any hints.
rt.representation-theory homological-algebra hochschild-homology
asked 2 hours ago
Lukas Woike
5387
add a comment |Â
up vote
6
down vote
favorite
Let $A$ be a finite-dimensional $k$-algebra and $U$ and $V$ two finite-dimensional projective $A$-modules (maybe neither the finiteness nor projectivity has to play a role, but these requirements are satisfied in my problem).
Now consider the $A$-bimodule
beginalign
M := mathrmHom_A (U,A) otimes mathrmHom_A (A,V) ,
endalign where $A$ is an $A$-module via left multiplication making $mathrmHom_A (U,A)$ an $A$-module and $mathrmHom_A (A,V)$ an $A^textopp$-module.
I would like to compute the Hochschild homology $HH_*(A;M)$.
If you write down the corresponding Hochschild complex, there is an augmentation by $mathrmHom_A (U,V)$. Does that induce a quasi-isomorphism?
Maybe such a result is known to experts?
Is there, more generally, a coefficient theorem I could use?
As usual, thank you for any hints.
rt.representation-theory homological-algebra hochschild-homology
asked 2 hours ago
Lukas Woike
5387
3
To get $operatornameHom_A(U,V)$ you do need f.g.p. of $U$, although nothing of $V$.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
2 hours ago
So the homology is $mathrmHom_A(U,V)$ (concentrated in degree zero) if $U$ is finitely generated projective? Can you provide an argument or a reference?
– Lukas Woike
2 hours ago
No no I don't know that much. I only wanted to say that the canonical map $operatornameHom_A(U,A)otimes_AVtooperatornameHom_A(U,V)$ is an isomorphism for all $V$ if and only if $U$ is fgp, nothing more
– áƒ›áƒÂმუკრჯიბლáƒÂძე
1 hour ago
Ok, I will see what I can do with that.
– Lukas Woike
1 hour ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $A$ be a finite-dimensional $k$-algebra and $U$ and $V$ two finite-dimensional projective $A$-modules (maybe neither the finiteness nor projectivity has to play a role, but these requirements are satisfied in my problem).
Now consider the $A$-bimodule
beginalign
M := mathrmHom_A (U,A) otimes mathrmHom_A (A,V) ,
endalign where $A$ is an $A$-module via left multiplication making $mathrmHom_A (U,A)$ an $A$-module and $mathrmHom_A (A,V)$ an $A^textopp$-module.
I would like to compute the Hochschild homology $HH_*(A;M)$.
If you write down the corresponding Hochschild complex, there is an augmentation by $mathrmHom_A (U,V)$. Does that induce a quasi-isomorphism?
Maybe such a result is known to experts?
Is there, more generally, a coefficient theorem I could use?
As usual, thank you for any hints.
rt.representation-theory homological-algebra hochschild-homology
asked 2 hours ago
Lukas Woike
5387
Let $A$ be a finite-dimensional $k$-algebra and $U$ and $V$ two finite-dimensional projective $A$-modules (maybe neither the finiteness nor projectivity has to play a role, but these requirements are satisfied in my problem).
Now consider the $A$-bimodule
beginalign
M := mathrmHom_A (U,A) otimes mathrmHom_A (A,V) ,
endalign where $A$ is an $A$-module via left multiplication making $mathrmHom_A (U,A)$ an $A$-module and $mathrmHom_A (A,V)$ an $A^textopp$-module.
I would like to compute the Hochschild homology $HH_*(A;M)$.
If you write down the corresponding Hochschild complex, there is an augmentation by $mathrmHom_A (U,V)$. Does that induce a quasi-isomorphism?
Maybe such a result is known to experts?
Is there, more generally, a coefficient theorem I could use?
As usual, thank you for any hints.
rt.representation-theory homological-algebra hochschild-homology
rt.representation-theory homological-algebra hochschild-homology
asked 2 hours ago
Lukas Woike
5387
asked 2 hours ago
Lukas Woike
5387
asked 2 hours ago
Lukas Woike
5387
asked 2 hours ago
Lukas Woike
5387
asked 2 hours ago
Lukas Woike
5387
5387
3
To get $operatornameHom_A(U,V)$ you do need f.g.p. of $U$, although nothing of $V$.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
2 hours ago
So the homology is $mathrmHom_A(U,V)$ (concentrated in degree zero) if $U$ is finitely generated projective? Can you provide an argument or a reference?
– Lukas Woike
2 hours ago
No no I don't know that much. I only wanted to say that the canonical map $operatornameHom_A(U,A)otimes_AVtooperatornameHom_A(U,V)$ is an isomorphism for all $V$ if and only if $U$ is fgp, nothing more
– áƒ›áƒÂმუკრჯიბლáƒÂძე
1 hour ago
Ok, I will see what I can do with that.
– Lukas Woike
1 hour ago
add a comment |Â
3
To get $operatornameHom_A(U,V)$ you do need f.g.p. of $U$, although nothing of $V$.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
2 hours ago
So the homology is $mathrmHom_A(U,V)$ (concentrated in degree zero) if $U$ is finitely generated projective? Can you provide an argument or a reference?
– Lukas Woike
2 hours ago
No no I don't know that much. I only wanted to say that the canonical map $operatornameHom_A(U,A)otimes_AVtooperatornameHom_A(U,V)$ is an isomorphism for all $V$ if and only if $U$ is fgp, nothing more
– áƒ›áƒÂმუკრჯიბლáƒÂძე
1 hour ago
Ok, I will see what I can do with that.
– Lukas Woike
1 hour ago
3
3
To get $operatornameHom_A(U,V)$ you do need f.g.p. of $U$, although nothing of $V$.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
2 hours ago
To get $operatornameHom_A(U,V)$ you do need f.g.p. of $U$, although nothing of $V$.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
2 hours ago
So the homology is $mathrmHom_A(U,V)$ (concentrated in degree zero) if $U$ is finitely generated projective? Can you provide an argument or a reference?
– Lukas Woike
2 hours ago
So the homology is $mathrmHom_A(U,V)$ (concentrated in degree zero) if $U$ is finitely generated projective? Can you provide an argument or a reference?
– Lukas Woike
2 hours ago
No no I don't know that much. I only wanted to say that the canonical map $operatornameHom_A(U,A)otimes_AVtooperatornameHom_A(U,V)$ is an isomorphism for all $V$ if and only if $U$ is fgp, nothing more
– áƒ›áƒÂმუკრჯიბლáƒÂძე
1 hour ago
No no I don't know that much. I only wanted to say that the canonical map $operatornameHom_A(U,A)otimes_AVtooperatornameHom_A(U,V)$ is an isomorphism for all $V$ if and only if $U$ is fgp, nothing more
– áƒ›áƒÂმუკრჯიბლáƒÂძე
1 hour ago
Ok, I will see what I can do with that.
– Lukas Woike
1 hour ago
Ok, I will see what I can do with that.
– Lukas Woike
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
(I'm assuming in the following that the base ring $k$ was a field.)
First, we note that for any $A$-bimodule of the form $M otimes N$, where $M$ is a left $A$-module and $N$ is a right $A$-module, has an isomorphism
$$
HH_*(A;M otimes N) cong Tor^A_*(N,M).
$$
To see this, we note that there is an explicit simplicial isomorphism between the cyclic bar construction computing Hochschild homology and the two-sided bar construction computing Tor:
$$
(M otimes N) otimes A^otimes p to N otimes A^otimes p otimes M
$$
In particular, the last map which moves the last factor of $A$ around and multiplies it on the left is carried simply to its left action on $M$.
Therefore, we find
$$
HH_*(A;Hom_A(U,A) otimes Hom_A(A,V)) cong Tor^A_*(Hom_A(A,V), Hom_A(U,A)) cong Tor^A_*(V, Hom_A(U,A)).
$$
If $V$ is projective as a (right) $A$-module, then we find that the Tor-groups vanish for $* > 0$ and that the zero'th group is
$$
V otimes_A Hom_A(U,A).
$$
The natural map augmentation that you are describing is then the natural map $V otimes_A Hom_A(U,A) to Hom_A(U,V)$, and (because $V$ is projective) this map is an isomorphism whenever $U$ is finitely presented. In particular, we don't need $U$ to be projective. (We could instead ask that $U$ is finitely generated projective and $V$ is arbitrary and get this result; this was stated in the comments already by მáƒÂმუკრჯიბლáƒÂძე.)
answered 1 hour ago
Tyler Lawson
38k7131196
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
(I'm assuming in the following that the base ring $k$ was a field.)
First, we note that for any $A$-bimodule of the form $M otimes N$, where $M$ is a left $A$-module and $N$ is a right $A$-module, has an isomorphism
$$
HH_*(A;M otimes N) cong Tor^A_*(N,M).
$$
To see this, we note that there is an explicit simplicial isomorphism between the cyclic bar construction computing Hochschild homology and the two-sided bar construction computing Tor:
$$
(M otimes N) otimes A^otimes p to N otimes A^otimes p otimes M
$$
In particular, the last map which moves the last factor of $A$ around and multiplies it on the left is carried simply to its left action on $M$.
Therefore, we find
$$
HH_*(A;Hom_A(U,A) otimes Hom_A(A,V)) cong Tor^A_*(Hom_A(A,V), Hom_A(U,A)) cong Tor^A_*(V, Hom_A(U,A)).
$$
If $V$ is projective as a (right) $A$-module, then we find that the Tor-groups vanish for $* > 0$ and that the zero'th group is
$$
V otimes_A Hom_A(U,A).
$$
The natural map augmentation that you are describing is then the natural map $V otimes_A Hom_A(U,A) to Hom_A(U,V)$, and (because $V$ is projective) this map is an isomorphism whenever $U$ is finitely presented. In particular, we don't need $U$ to be projective. (We could instead ask that $U$ is finitely generated projective and $V$ is arbitrary and get this result; this was stated in the comments already by მáƒÂმუკრჯიბლáƒÂძე.)
answered 1 hour ago
Tyler Lawson
38k7131196
add a comment |Â
up vote
3
down vote
(I'm assuming in the following that the base ring $k$ was a field.)
First, we note that for any $A$-bimodule of the form $M otimes N$, where $M$ is a left $A$-module and $N$ is a right $A$-module, has an isomorphism
$$
HH_*(A;M otimes N) cong Tor^A_*(N,M).
$$
To see this, we note that there is an explicit simplicial isomorphism between the cyclic bar construction computing Hochschild homology and the two-sided bar construction computing Tor:
$$
(M otimes N) otimes A^otimes p to N otimes A^otimes p otimes M
$$
In particular, the last map which moves the last factor of $A$ around and multiplies it on the left is carried simply to its left action on $M$.
Therefore, we find
$$
HH_*(A;Hom_A(U,A) otimes Hom_A(A,V)) cong Tor^A_*(Hom_A(A,V), Hom_A(U,A)) cong Tor^A_*(V, Hom_A(U,A)).
$$
If $V$ is projective as a (right) $A$-module, then we find that the Tor-groups vanish for $* > 0$ and that the zero'th group is
$$
V otimes_A Hom_A(U,A).
$$
The natural map augmentation that you are describing is then the natural map $V otimes_A Hom_A(U,A) to Hom_A(U,V)$, and (because $V$ is projective) this map is an isomorphism whenever $U$ is finitely presented. In particular, we don't need $U$ to be projective. (We could instead ask that $U$ is finitely generated projective and $V$ is arbitrary and get this result; this was stated in the comments already by მáƒÂმუკრჯიბლáƒÂძე.)
answered 1 hour ago
Tyler Lawson
38k7131196
add a comment |Â
up vote
3
down vote
up vote
3
down vote
(I'm assuming in the following that the base ring $k$ was a field.)
First, we note that for any $A$-bimodule of the form $M otimes N$, where $M$ is a left $A$-module and $N$ is a right $A$-module, has an isomorphism
$$
HH_*(A;M otimes N) cong Tor^A_*(N,M).
$$
To see this, we note that there is an explicit simplicial isomorphism between the cyclic bar construction computing Hochschild homology and the two-sided bar construction computing Tor:
$$
(M otimes N) otimes A^otimes p to N otimes A^otimes p otimes M
$$
In particular, the last map which moves the last factor of $A$ around and multiplies it on the left is carried simply to its left action on $M$.
Therefore, we find
$$
HH_*(A;Hom_A(U,A) otimes Hom_A(A,V)) cong Tor^A_*(Hom_A(A,V), Hom_A(U,A)) cong Tor^A_*(V, Hom_A(U,A)).
$$
If $V$ is projective as a (right) $A$-module, then we find that the Tor-groups vanish for $* > 0$ and that the zero'th group is
$$
V otimes_A Hom_A(U,A).
$$
The natural map augmentation that you are describing is then the natural map $V otimes_A Hom_A(U,A) to Hom_A(U,V)$, and (because $V$ is projective) this map is an isomorphism whenever $U$ is finitely presented. In particular, we don't need $U$ to be projective. (We could instead ask that $U$ is finitely generated projective and $V$ is arbitrary and get this result; this was stated in the comments already by მáƒÂმუკრჯიბლáƒÂძე.)
answered 1 hour ago
Tyler Lawson
38k7131196
(I'm assuming in the following that the base ring $k$ was a field.)
First, we note that for any $A$-bimodule of the form $M otimes N$, where $M$ is a left $A$-module and $N$ is a right $A$-module, has an isomorphism
$$
HH_*(A;M otimes N) cong Tor^A_*(N,M).
$$
To see this, we note that there is an explicit simplicial isomorphism between the cyclic bar construction computing Hochschild homology and the two-sided bar construction computing Tor:
$$
(M otimes N) otimes A^otimes p to N otimes A^otimes p otimes M
$$
In particular, the last map which moves the last factor of $A$ around and multiplies it on the left is carried simply to its left action on $M$.
Therefore, we find
$$
HH_*(A;Hom_A(U,A) otimes Hom_A(A,V)) cong Tor^A_*(Hom_A(A,V), Hom_A(U,A)) cong Tor^A_*(V, Hom_A(U,A)).
$$
If $V$ is projective as a (right) $A$-module, then we find that the Tor-groups vanish for $* > 0$ and that the zero'th group is
$$
V otimes_A Hom_A(U,A).
$$
The natural map augmentation that you are describing is then the natural map $V otimes_A Hom_A(U,A) to Hom_A(U,V)$, and (because $V$ is projective) this map is an isomorphism whenever $U$ is finitely presented. In particular, we don't need $U$ to be projective. (We could instead ask that $U$ is finitely generated projective and $V$ is arbitrary and get this result; this was stated in the comments already by მáƒÂმუკრჯიბლáƒÂძე.)
answered 1 hour ago
Tyler Lawson
38k7131196
answered 1 hour ago
Tyler Lawson
38k7131196
answered 1 hour ago
Tyler Lawson
38k7131196
answered 1 hour ago
Tyler Lawson
38k7131196
38k7131196
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f314045%2fhochschild-homology-with-coefficients-in-a-certain-bimodule%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
To get $operatornameHom_A(U,V)$ you do need f.g.p. of $U$, although nothing of $V$.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
2 hours ago
So the homology is $mathrmHom_A(U,V)$ (concentrated in degree zero) if $U$ is finitely generated projective? Can you provide an argument or a reference?
– Lukas Woike
2 hours ago
No no I don't know that much. I only wanted to say that the canonical map $operatornameHom_A(U,A)otimes_AVtooperatornameHom_A(U,V)$ is an isomorphism for all $V$ if and only if $U$ is fgp, nothing more
– áƒ›áƒÂმუკრჯიბლáƒÂძე
1 hour ago
Ok, I will see what I can do with that.
– Lukas Woike
1 hour ago