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Background: Young symmetrizer $c_\lambda$ gives an explicit description of Schur module $S_\lambda V$, which is also the kernel of maps between exterior products (as in Fulton & Harris).

Example: Consider the Young symmetrizer $c_{(21)} = () + (12) - (13) - (132)$, constructed from:

$1 2\\3$

Then its image in $\otimes^3 V$ is spanned by:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

which also gives $S_{(2,1)}V$. And we have:

$S_{(2,1)}V = Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

=======================

Question 1: The dimension of $S_{(2,1)}V$ shall be $\frac{(n+1)n(n-1)}{3}$, where $n = dim(V)$. However, I am not sure how to directly see the expression:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

has dimension $\frac{(n+1)n(n-1)}{3}$ in $\otimes^3 V$.

That is, without using $\dim( S_{(2,1)}) = \dim(\Lambda^2V \otimes V) - \dim(\Lambda^3V)$.

=======================

Question 2: Consider the map $\Lambda^2V \otimes V \to \Lambda^3V$ again, which is defined by:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1 \wedge v_2 \wedge v_3$

It seems there are multiple ways to write down its kernel, such as:

$(v_1 \wedge v_2) \otimes v_3 + (v_1 \wedge v_3) \otimes v_2$, or,

$(v_1 \wedge v_3) \otimes v_2 + (v_2 \wedge v_3) \otimes v_1$, etc.

And there are multiple ways to construct a map $\Lambda^2V \otimes V \to \otimes^3 V$, such as:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_2 \otimes v_3 - v_2 \otimes v_1 \otimes v_3$, or,

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_3 \otimes v_2 - v_2 \otimes v_3 \otimes v_1$, etc.

So there are actually multiple elements in $\otimes^3 V$ corresponding to $Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

I have computed some of them, and it seems they are either:

  • Young symmetrizer $c^\prime_{(21)}$ constructed from all ways of filling the Young tableaux, such as $() + (23) - (12) - (132)$ constructed from:

$2 3\\1$

  • $g \cdot c^\prime_{(21)}$ where $g \in S_3$. Example: $(23) + (123) - (132) - (13) = (23) \cdot \Big(() + (13) - (12) - (123)\Big)$. And $() + (13) - (12) - (123)$ can be constructed from:

$1 3\\2$

So my second questions is, is the above true for other partitions $\lambda$?

Background: Young symmetrizer $c_\lambda$ gives an explicit description of Schur module $S_\lambda V$, which is also the kernel of maps between exterior products (as in Fulton & Harris).

Example: Consider the Young symmetrizer $c_{(21)} = () + (12) - (13) - (132)$, constructed from:

$1 2\\3$

Then its image in $\otimes^3 V$ is spanned by:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

which also gives $S_{(2,1)}V$. And we have:

$S_{(2,1)}V = Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

=======================

Question 1: The dimension of $S_{(2,1)}V$ shall be $\frac{(n+1)n(n-1)}{3}$, where $n = dim(V)$. However, I am not sure how to see:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

has dimension $\frac{(n+1)n(n-1)}{3}$ in $\otimes^3 V$.

=======================

Question 2: Consider the map $\Lambda^2V \otimes V \to \Lambda^3V$ again, which is defined by:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1 \wedge v_2 \wedge v_3$

It seems there are multiple ways to write down its kernel, such as:

$(v_1 \wedge v_2) \otimes v_3 + (v_1 \wedge v_3) \otimes v_2$, or,

$(v_1 \wedge v_3) \otimes v_2 + (v_2 \wedge v_3) \otimes v_1$, etc.

And there are multiple ways to construct a map $\Lambda^2V \otimes V \to \otimes^3 V$, such as:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_2 \otimes v_3 - v_2 \otimes v_1 \otimes v_3$, or,

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_3 \otimes v_2 - v_2 \otimes v_3 \otimes v_1$, etc.

So there are actually multiple elements in $\otimes^3 V$ corresponding to $Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

I have computed some of them, and it seems they are either:

  • Young symmetrizer $c^\prime_{(21)}$ constructed from all ways of filling the Young tableaux, such as $() + (23) - (12) - (132)$ constructed from:

$2 3\\1$

  • $g \cdot c^\prime_{(21)}$ where $g \in S_3$. Example: $(23) + (123) - (132) - (13) = (23) \cdot \Big(() + (13) - (12) - (123)\Big)$.

So my second questions is, is the above true for other partitions $\lambda$?

Background: Young symmetrizer $c_\lambda$ gives an explicit description of Schur module $S_\lambda V$, which is also the kernel of maps between exterior products (as in Fulton & Harris).

Example: Consider the Young symmetrizer $c_{(21)} = () + (12) - (13) - (132)$, constructed from:

$1 2\\3$

Then its image in $\otimes^3 V$ is spanned by:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

which also gives $S_{(2,1)}V$. And we have:

$S_{(2,1)}V = Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

=======================

Question 1: The dimension of $S_{(2,1)}V$ shall be $\frac{(n+1)n(n-1)}{3}$, where $n = dim(V)$. However, I am not sure how to directly see the expression:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

has dimension $\frac{(n+1)n(n-1)}{3}$ in $\otimes^3 V$.

That is, without using $\dim( S_{(2,1)}) = \dim(\Lambda^2V \otimes V) - \dim(\Lambda^3V)$.

=======================

Question 2: Consider the map $\Lambda^2V \otimes V \to \Lambda^3V$ again, which is defined by:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1 \wedge v_2 \wedge v_3$

It seems there are multiple ways to write down its kernel, such as:

$(v_1 \wedge v_2) \otimes v_3 + (v_1 \wedge v_3) \otimes v_2$, or,

$(v_1 \wedge v_3) \otimes v_2 + (v_2 \wedge v_3) \otimes v_1$, etc.

And there are multiple ways to construct a map $\Lambda^2V \otimes V \to \otimes^3 V$, such as:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_2 \otimes v_3 - v_2 \otimes v_1 \otimes v_3$, or,

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_3 \otimes v_2 - v_2 \otimes v_3 \otimes v_1$, etc.

So there are actually multiple elements in $\otimes^3 V$ corresponding to $Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

I have computed some of them, and it seems they are either:

  • Young symmetrizer $c^\prime_{(21)}$ constructed from all ways of filling the Young tableaux, such as $() + (23) - (12) - (132)$ constructed from:

$2 3\\1$

  • $g \cdot c^\prime_{(21)}$ where $g \in S_3$. Example: $(23) + (123) - (132) - (13) = (23) \cdot \Big(() + (13) - (12) - (123)\Big)$. And $() + (13) - (12) - (123)$ can be constructed from:

$1 3\\2$

So my second questions is, is the above true for other partitions $\lambda$?

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Young Symmetrizer and Exterior Products, such as $S_{(2,1)}V = Ker(\Lambda^2V \otimes V \to \Lambda^3V )$

Background: Young symmetrizer $c_\lambda$ gives an explicit description of Schur module $S_\lambda V$, which is also the kernel of maps between exterior products (as in Fulton & Harris).

Example: Consider the Young symmetrizer $c_{(21)} = () + (12) - (13) - (132)$, constructed from:

$1 2\\3$

Then its image in $\otimes^3 V$ is spanned by:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

which also gives $S_{(2,1)}V$. And we have:

$S_{(2,1)}V = Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

=======================

Question 1: The dimension of $S_{(2,1)}V$ shall be $\frac{(n+1)n(n-1)}{3}$, where $n = dim(V)$. However, I am not sure how to see:

$v_1 \otimes v_2 \otimes v_3 + v_2 \otimes v_1 \otimes v_3 - v_3 \otimes v_2 \otimes v_1 - v_3 \otimes v_1 \otimes v_2$

has dimension $\frac{(n+1)n(n-1)}{3}$ in $\otimes^3 V$.

=======================

Question 2: Consider the map $\Lambda^2V \otimes V \to \Lambda^3V$ again, which is defined by:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1 \wedge v_2 \wedge v_3$

It seems there are multiple ways to write down its kernel, such as:

$(v_1 \wedge v_2) \otimes v_3 + (v_1 \wedge v_3) \otimes v_2$, or,

$(v_1 \wedge v_3) \otimes v_2 + (v_2 \wedge v_3) \otimes v_1$, etc.

And there are multiple ways to construct a map $\Lambda^2V \otimes V \to \otimes^3 V$, such as:

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_2 \otimes v_3 - v_2 \otimes v_1 \otimes v_3$, or,

$(v_1 \wedge v_2) \otimes v_3 \mapsto v_1\otimes v_3 \otimes v_2 - v_2 \otimes v_3 \otimes v_1$, etc.

So there are actually multiple elements in $\otimes^3 V$ corresponding to $Ker(\Lambda^2V \otimes V \to \Lambda^3V )$.

I have computed some of them, and it seems they are either:

  • Young symmetrizer $c^\prime_{(21)}$ constructed from all ways of filling the Young tableaux, such as $() + (23) - (12) - (132)$ constructed from:

$2 3\\1$

  • $g \cdot c^\prime_{(21)}$ where $g \in S_3$. Example: $(23) + (123) - (132) - (13) = (23) \cdot \Big(() + (13) - (12) - (123)\Big)$.

So my second questions is, is the above true for other partitions $\lambda$?