Hom and Tensoring with the Dual in Category O
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In general, for infinite dimensional vector spaces, we know that $V^* otimes W$ and $text{Hom}(V,W)$ are not isomorphic, with $V^* otimes W$ being isomorphic instead to the subspace of finite rank maps.
Now, suppose instead that $V$, $W$ are modules in Category $mathcal{O}$ for some semi-simple Lie algebra (or, more generally a symmetrizable Kac-Moody algebra). If, instead of the full dual $V^* = prod_lambda V^*_lambda$, we use the restricted dual $V^vee := bigoplus_lambda V^*_lambda$ (to stay inside of $mathcal{O}$), do we recover a statement like $V^vee otimes W cong text{Hom}(V,W)$?
representation-theory lie-algebras
asked Dec 10 '18 at 3:33
SamJeraldsSamJeralds
261
$endgroup$
add a comment |
$begingroup$
In general, for infinite dimensional vector spaces, we know that $V^* otimes W$ and $text{Hom}(V,W)$ are not isomorphic, with $V^* otimes W$ being isomorphic instead to the subspace of finite rank maps.
Now, suppose instead that $V$, $W$ are modules in Category $mathcal{O}$ for some semi-simple Lie algebra (or, more generally a symmetrizable Kac-Moody algebra). If, instead of the full dual $V^* = prod_lambda V^*_lambda$, we use the restricted dual $V^vee := bigoplus_lambda V^*_lambda$ (to stay inside of $mathcal{O}$), do we recover a statement like $V^vee otimes W cong text{Hom}(V,W)$?
representation-theory lie-algebras
asked Dec 10 '18 at 3:33
SamJeraldsSamJeralds
261
$endgroup$
1
$begingroup$
You need to be careful. In order to take this dual (which I would usually prefer to call the graded dual rather than the restricted dual, since it really does depend on a grading of the underlying vector space), you also need to "twist" the action to get back into category $mathcal{O}$. But it will still not work quite this nicely, since the Hom can still recover the "usual" dual by taking $W$ to be the trivial module, in which case the dimensions don't match.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:27
$begingroup$
I guess the obvious replacement for Hom would be $operatorname{Hom}(V, W) := bigoplus_lambda operatorname{Hom}_{mathbb{C}}(V_lambda, W_lambda)$, and this would be isomorphic to $V^vee otimes W$ as $mathbb{C}$-vector spaces. However, note that it's pretty rare that the tensor product of objects in $mathcal{O}$ is again in $mathcal{O}$.
$endgroup$
– Joppy
Dec 10 '18 at 12:53
add a comment |
$begingroup$
In general, for infinite dimensional vector spaces, we know that $V^* otimes W$ and $text{Hom}(V,W)$ are not isomorphic, with $V^* otimes W$ being isomorphic instead to the subspace of finite rank maps.
Now, suppose instead that $V$, $W$ are modules in Category $mathcal{O}$ for some semi-simple Lie algebra (or, more generally a symmetrizable Kac-Moody algebra). If, instead of the full dual $V^* = prod_lambda V^*_lambda$, we use the restricted dual $V^vee := bigoplus_lambda V^*_lambda$ (to stay inside of $mathcal{O}$), do we recover a statement like $V^vee otimes W cong text{Hom}(V,W)$?
representation-theory lie-algebras
asked Dec 10 '18 at 3:33
SamJeraldsSamJeralds
261
$endgroup$
In general, for infinite dimensional vector spaces, we know that $V^* otimes W$ and $text{Hom}(V,W)$ are not isomorphic, with $V^* otimes W$ being isomorphic instead to the subspace of finite rank maps.
Now, suppose instead that $V$, $W$ are modules in Category $mathcal{O}$ for some semi-simple Lie algebra (or, more generally a symmetrizable Kac-Moody algebra). If, instead of the full dual $V^* = prod_lambda V^*_lambda$, we use the restricted dual $V^vee := bigoplus_lambda V^*_lambda$ (to stay inside of $mathcal{O}$), do we recover a statement like $V^vee otimes W cong text{Hom}(V,W)$?
representation-theory lie-algebras
representation-theory lie-algebras
asked Dec 10 '18 at 3:33
SamJeraldsSamJeralds
261
asked Dec 10 '18 at 3:33
SamJeraldsSamJeralds
261
asked Dec 10 '18 at 3:33
SamJeraldsSamJeralds
261
asked Dec 10 '18 at 3:33
SamJeraldsSamJeralds
261
asked Dec 10 '18 at 3:33
SamJeraldsSamJeralds
261
261
1
$begingroup$
You need to be careful. In order to take this dual (which I would usually prefer to call the graded dual rather than the restricted dual, since it really does depend on a grading of the underlying vector space), you also need to "twist" the action to get back into category $mathcal{O}$. But it will still not work quite this nicely, since the Hom can still recover the "usual" dual by taking $W$ to be the trivial module, in which case the dimensions don't match.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:27
$begingroup$
I guess the obvious replacement for Hom would be $operatorname{Hom}(V, W) := bigoplus_lambda operatorname{Hom}_{mathbb{C}}(V_lambda, W_lambda)$, and this would be isomorphic to $V^vee otimes W$ as $mathbb{C}$-vector spaces. However, note that it's pretty rare that the tensor product of objects in $mathcal{O}$ is again in $mathcal{O}$.
$endgroup$
– Joppy
Dec 10 '18 at 12:53
add a comment |
1
$begingroup$
You need to be careful. In order to take this dual (which I would usually prefer to call the graded dual rather than the restricted dual, since it really does depend on a grading of the underlying vector space), you also need to "twist" the action to get back into category $mathcal{O}$. But it will still not work quite this nicely, since the Hom can still recover the "usual" dual by taking $W$ to be the trivial module, in which case the dimensions don't match.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:27
$begingroup$
I guess the obvious replacement for Hom would be $operatorname{Hom}(V, W) := bigoplus_lambda operatorname{Hom}_{mathbb{C}}(V_lambda, W_lambda)$, and this would be isomorphic to $V^vee otimes W$ as $mathbb{C}$-vector spaces. However, note that it's pretty rare that the tensor product of objects in $mathcal{O}$ is again in $mathcal{O}$.
$endgroup$
– Joppy
Dec 10 '18 at 12:53
1
1
$begingroup$
You need to be careful. In order to take this dual (which I would usually prefer to call the graded dual rather than the restricted dual, since it really does depend on a grading of the underlying vector space), you also need to "twist" the action to get back into category $mathcal{O}$. But it will still not work quite this nicely, since the Hom can still recover the "usual" dual by taking $W$ to be the trivial module, in which case the dimensions don't match.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:27
$begingroup$
You need to be careful. In order to take this dual (which I would usually prefer to call the graded dual rather than the restricted dual, since it really does depend on a grading of the underlying vector space), you also need to "twist" the action to get back into category $mathcal{O}$. But it will still not work quite this nicely, since the Hom can still recover the "usual" dual by taking $W$ to be the trivial module, in which case the dimensions don't match.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:27
$begingroup$
I guess the obvious replacement for Hom would be $operatorname{Hom}(V, W) := bigoplus_lambda operatorname{Hom}_{mathbb{C}}(V_lambda, W_lambda)$, and this would be isomorphic to $V^vee otimes W$ as $mathbb{C}$-vector spaces. However, note that it's pretty rare that the tensor product of objects in $mathcal{O}$ is again in $mathcal{O}$.
$endgroup$
– Joppy
Dec 10 '18 at 12:53
$begingroup$
I guess the obvious replacement for Hom would be $operatorname{Hom}(V, W) := bigoplus_lambda operatorname{Hom}_{mathbb{C}}(V_lambda, W_lambda)$, and this would be isomorphic to $V^vee otimes W$ as $mathbb{C}$-vector spaces. However, note that it's pretty rare that the tensor product of objects in $mathcal{O}$ is again in $mathcal{O}$.
$endgroup$
– Joppy
Dec 10 '18 at 12:53
add a comment |
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$begingroup$
You need to be careful. In order to take this dual (which I would usually prefer to call the graded dual rather than the restricted dual, since it really does depend on a grading of the underlying vector space), you also need to "twist" the action to get back into category $mathcal{O}$. But it will still not work quite this nicely, since the Hom can still recover the "usual" dual by taking $W$ to be the trivial module, in which case the dimensions don't match.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:27
$begingroup$
I guess the obvious replacement for Hom would be $operatorname{Hom}(V, W) := bigoplus_lambda operatorname{Hom}_{mathbb{C}}(V_lambda, W_lambda)$, and this would be isomorphic to $V^vee otimes W$ as $mathbb{C}$-vector spaces. However, note that it's pretty rare that the tensor product of objects in $mathcal{O}$ is again in $mathcal{O}$.
$endgroup$
– Joppy
Dec 10 '18 at 12:53