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Post Closed as "Not suitable for this site" by Vladimir Dotsenko, Stefan Waldmann, abx, Marco Golla, Karl Schwede
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Jianrong Li
Jianrong Li
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Let $V, W$ be two vector spaces. We have $\Lambda^2(V \otimes W) = (\Lambda^2 V \otimes S^2 V) \oplus (S^2 V \otimes \Lambda^2 V)$$\Lambda^2(V \otimes W) \cong (\Lambda^2 V \otimes S^2 W) \oplus (S^2 V \otimes \Lambda^2 W)$. I am trying to find similar results for $\Lambda^3(V \otimes W)$. I think that $\Lambda^3 V \otimes S^3 V$$\Lambda^3 V \otimes S^3 W$ and $S^3 V \otimes \Lambda^3 V$$S^3 V \otimes \Lambda^3 W$ are subspaces of $\Lambda^3(V \otimes W)$. It seems that $\Lambda^3(V \otimes W)$ has other subspaces since if the dimensions of $V$ and $W$ are $n, m$ respectively, then the dimension of $\Lambda^3(V \otimes W)$ is $\frac{m\, n\, \left(m\, n - 1\right)\, \left(m\, n - 2\right)}{6}$ and the sum of dimensions of $\Lambda^3 V \otimes S^3 V$$\Lambda^3 V \otimes S^3 W$ and $S^3 V \otimes \Lambda^3 V$$S^3 V \otimes \Lambda^3 W$ is $\frac{m\, n\, \left(m^2\, n^2 + 2\, m^2 - 9\, m\, n + 2\, n^2 + 4\right)}{18}$. The difference of these two numbers is $\frac{m\, n\, \left(n - 1\right)\, \left(n + 1\right)\, \left(m - 1\right)\, \left(m + 1\right)}{9}$. Therefore we have $$ \Lambda^3(V \otimes W) = (\Lambda^3 V \otimes S^3 V) \oplus (S^3 V \otimes \Lambda^3 V) \oplus W_1. $$$$ \Lambda^3(V \otimes W) \cong (\Lambda^3 V \otimes S^3 W) \oplus (S^3 V \otimes \Lambda^3 W) \oplus W_1. $$ How to compute $W_1$ explicitly? Any help will be greatly appreciated!

Let $V, W$ be two vector spaces. We have $\Lambda^2(V \otimes W) = (\Lambda^2 V \otimes S^2 V) \oplus (S^2 V \otimes \Lambda^2 V)$. I am trying to find similar results for $\Lambda^3(V \otimes W)$. I think that $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ are subspaces of $\Lambda^3(V \otimes W)$. It seems that $\Lambda^3(V \otimes W)$ has other subspaces since if the dimensions of $V$ and $W$ are $n, m$ respectively, then the dimension of $\Lambda^3(V \otimes W)$ is $\frac{m\, n\, \left(m\, n - 1\right)\, \left(m\, n - 2\right)}{6}$ and the sum of dimensions of $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ is $\frac{m\, n\, \left(m^2\, n^2 + 2\, m^2 - 9\, m\, n + 2\, n^2 + 4\right)}{18}$. The difference of these two numbers is $\frac{m\, n\, \left(n - 1\right)\, \left(n + 1\right)\, \left(m - 1\right)\, \left(m + 1\right)}{9}$. Therefore we have $$ \Lambda^3(V \otimes W) = (\Lambda^3 V \otimes S^3 V) \oplus (S^3 V \otimes \Lambda^3 V) \oplus W_1. $$ How to compute $W_1$ explicitly? Any help will be greatly appreciated!

Let $V, W$ be two vector spaces. We have $\Lambda^2(V \otimes W) \cong (\Lambda^2 V \otimes S^2 W) \oplus (S^2 V \otimes \Lambda^2 W)$. I am trying to find similar results for $\Lambda^3(V \otimes W)$. I think that $\Lambda^3 V \otimes S^3 W$ and $S^3 V \otimes \Lambda^3 W$ are subspaces of $\Lambda^3(V \otimes W)$. It seems that $\Lambda^3(V \otimes W)$ has other subspaces since if the dimensions of $V$ and $W$ are $n, m$ respectively, then the dimension of $\Lambda^3(V \otimes W)$ is $\frac{m\, n\, \left(m\, n - 1\right)\, \left(m\, n - 2\right)}{6}$ and the sum of dimensions of $\Lambda^3 V \otimes S^3 W$ and $S^3 V \otimes \Lambda^3 W$ is $\frac{m\, n\, \left(m^2\, n^2 + 2\, m^2 - 9\, m\, n + 2\, n^2 + 4\right)}{18}$. The difference of these two numbers is $\frac{m\, n\, \left(n - 1\right)\, \left(n + 1\right)\, \left(m - 1\right)\, \left(m + 1\right)}{9}$. Therefore we have $$ \Lambda^3(V \otimes W) \cong (\Lambda^3 V \otimes S^3 W) \oplus (S^3 V \otimes \Lambda^3 W) \oplus W_1. $$ How to compute $W_1$ explicitly? Any help will be greatly appreciated!

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Jianrong Li
Jianrong Li
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Let $V, W$ be two vector spaces. We have $\Lambda^2(V \otimes W) = \Lambda^2 V \otimes S^2 V \oplus S^2 V \otimes \Lambda^2 V$$\Lambda^2(V \otimes W) = (\Lambda^2 V \otimes S^2 V) \oplus (S^2 V \otimes \Lambda^2 V)$. I am trying to find similar results for $\Lambda^3(V \otimes W)$. I think that $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ are subspaces of $\Lambda^3(V \otimes W)$. It seems that $\Lambda^3(V \otimes W)$ has other subspaces since if the dimensions of $V$ and $W$ are $n, m$ respectively, then the dimension of $\Lambda^3(V \otimes W)$ is $\frac{m\, n\, \left(m\, n - 1\right)\, \left(m\, n - 2\right)}{6}$ and the sum of dimensions of $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ is $\frac{m\, n\, \left(m^2\, n^2 + 2\, m^2 - 9\, m\, n + 2\, n^2 + 4\right)}{18}$. The difference of these two numbers is $\frac{m\, n\, \left(n - 1\right)\, \left(n + 1\right)\, \left(m - 1\right)\, \left(m + 1\right)}{9}$. Therefore we have $$ \Lambda^3(V \otimes W) = \Lambda^3 V \otimes S^3 V \oplus S^3 V \otimes \Lambda^3 V \oplus W_1. $$$$ \Lambda^3(V \otimes W) = (\Lambda^3 V \otimes S^3 V) \oplus (S^3 V \otimes \Lambda^3 V) \oplus W_1. $$ How to compute $W_1$ explicitly? Any help will be greatly appreciated!

Let $V, W$ be two vector spaces. We have $\Lambda^2(V \otimes W) = \Lambda^2 V \otimes S^2 V \oplus S^2 V \otimes \Lambda^2 V$. I am trying to find similar results for $\Lambda^3(V \otimes W)$. I think that $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ are subspaces of $\Lambda^3(V \otimes W)$. It seems that $\Lambda^3(V \otimes W)$ has other subspaces since if the dimensions of $V$ and $W$ are $n, m$ respectively, then the dimension of $\Lambda^3(V \otimes W)$ is $\frac{m\, n\, \left(m\, n - 1\right)\, \left(m\, n - 2\right)}{6}$ and the sum of dimensions of $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ is $\frac{m\, n\, \left(m^2\, n^2 + 2\, m^2 - 9\, m\, n + 2\, n^2 + 4\right)}{18}$. The difference of these two numbers is $\frac{m\, n\, \left(n - 1\right)\, \left(n + 1\right)\, \left(m - 1\right)\, \left(m + 1\right)}{9}$. Therefore we have $$ \Lambda^3(V \otimes W) = \Lambda^3 V \otimes S^3 V \oplus S^3 V \otimes \Lambda^3 V \oplus W_1. $$ How to compute $W_1$ explicitly? Any help will be greatly appreciated!

Let $V, W$ be two vector spaces. We have $\Lambda^2(V \otimes W) = (\Lambda^2 V \otimes S^2 V) \oplus (S^2 V \otimes \Lambda^2 V)$. I am trying to find similar results for $\Lambda^3(V \otimes W)$. I think that $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ are subspaces of $\Lambda^3(V \otimes W)$. It seems that $\Lambda^3(V \otimes W)$ has other subspaces since if the dimensions of $V$ and $W$ are $n, m$ respectively, then the dimension of $\Lambda^3(V \otimes W)$ is $\frac{m\, n\, \left(m\, n - 1\right)\, \left(m\, n - 2\right)}{6}$ and the sum of dimensions of $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ is $\frac{m\, n\, \left(m^2\, n^2 + 2\, m^2 - 9\, m\, n + 2\, n^2 + 4\right)}{18}$. The difference of these two numbers is $\frac{m\, n\, \left(n - 1\right)\, \left(n + 1\right)\, \left(m - 1\right)\, \left(m + 1\right)}{9}$. Therefore we have $$ \Lambda^3(V \otimes W) = (\Lambda^3 V \otimes S^3 V) \oplus (S^3 V \otimes \Lambda^3 V) \oplus W_1. $$ How to compute $W_1$ explicitly? Any help will be greatly appreciated!

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Jianrong Li
Jianrong Li
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Decompose $\Lambda^3(V \otimes W)$

Let $V, W$ be two vector spaces. We have $\Lambda^2(V \otimes W) = \Lambda^2 V \otimes S^2 V \oplus S^2 V \otimes \Lambda^2 V$. I am trying to find similar results for $\Lambda^3(V \otimes W)$. I think that $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ are subspaces of $\Lambda^3(V \otimes W)$. It seems that $\Lambda^3(V \otimes W)$ has other subspaces since if the dimensions of $V$ and $W$ are $n, m$ respectively, then the dimension of $\Lambda^3(V \otimes W)$ is $\frac{m\, n\, \left(m\, n - 1\right)\, \left(m\, n - 2\right)}{6}$ and the sum of dimensions of $\Lambda^3 V \otimes S^3 V$ and $S^3 V \otimes \Lambda^3 V$ is $\frac{m\, n\, \left(m^2\, n^2 + 2\, m^2 - 9\, m\, n + 2\, n^2 + 4\right)}{18}$. The difference of these two numbers is $\frac{m\, n\, \left(n - 1\right)\, \left(n + 1\right)\, \left(m - 1\right)\, \left(m + 1\right)}{9}$. Therefore we have $$ \Lambda^3(V \otimes W) = \Lambda^3 V \otimes S^3 V \oplus S^3 V \otimes \Lambda^3 V \oplus W_1. $$ How to compute $W_1$ explicitly? Any help will be greatly appreciated!