5
$\begingroup$

I have asked this question in MSE before, but have not got any answer. So here I am asking it again with some more detail.

I believe that the following sequence of $\mathfrak{sl}_3(\mathbb{C})$-modules is exact:

$$0\to\mathbb{C}\to V\otimes\wedge^2V\stackrel\phi\to S^2 V\otimes S^2(\wedge^2 V)\stackrel\psi\to S^3(S^2V)\stackrel{N}\to S^6V\to 0$$

where $V=\mathbb{C}^3$ and $S^2$ denotes the symmetric square. The exactness of this sequence can be shown easily using eigen value diagram. But I want to show the exactness explicitly by constructing the maps and showing exactness of those maps.

Let's call the third and fourth map $\phi$ and $\psi$ respectively, the fifth map is the natural map (call it $N$). I also know what $\psi$ is, it is given by the followig $$\psi ((s\cdot t)\otimes (u\wedge v)\cdot(w\wedge z))=((u\cdot w)\cdot(v\cdot z)-(u\cdot z)\cdot(v\cdot w))\cdot (s\cdot t)$$ and I have checked that $\operatorname{im}\psi=\ker N$. But I am not sure what $\phi$ is.

So my question is, how to define $\phi$ so that $\ker\phi$ is isomorphic to $\mathbb{C}$ and $\psi\circ\phi$ is zero i.e. $\operatorname{im}\phi\subset\ker\psi$ (the other direction $\ker\psi\subset\operatorname{im}\phi$ follows by computing the dimensions which turns out to be equal) and of course $\phi$ should be $\mathfrak{sl}_3(\mathbb{C})$-linear.

$\endgroup$

1 Answer 1

6
$\begingroup$

Under the action of $\mathfrak{sl}_3(\mathbb{C})$, $\Lambda^2V\simeq V^\ast$. This isomorphism can be written explicitly in terms of the invariant volume form $\omega\in\Lambda^3V\simeq \mathbb{C} $, by $(v_1,v_2\wedge v_3)=v_1\wedge v_2\wedge v_3=\lambda \omega$ (in other words, for $\theta\in V^\ast$, $\theta\mapsto \iota_\theta\omega\in\Lambda^2V$).

Thus $V\otimes\Lambda^2V\simeq V\otimes V^\ast = \mathfrak{gl}_3(\mathbb{C}) = \mathfrak{sl}_3(\mathbb{C})\oplus \mathbb{C}$. Moreover, $S^2V\otimes S^2(\Lambda^2 V)\simeq S^2V\otimes S^2 V^\ast = \mathfrak{gl}(S^2V)$. The latter space contains a subalgebra isomorphic to $\mathfrak{sl}_3(\mathbb{C})$, which is spanned by the representation matrices of the induced action on $S^2V$.

Your map $\phi$ is then given by composing the following operations: first use the isomorphism $\Lambda^2V\simeq V^\ast$, then remove trace of the resulting operator by $A\mapsto A-\text{Tr}(A)/3 I$, then send the resulting matrix to its induced matrix on $S^2V$, and finally use the isomorphism $\Lambda^2V\simeq V^\ast$ to obtain an element of $S^2V\otimes S^2(\Lambda^2 V)$.

The kernel property is clear, because $S^3S^2V$ contains no submodule isomorpic to $\mathfrak{sl}_3(\mathbb{C})$. (This follows from highest weight calculations).

$\endgroup$
1
  • $\begingroup$ Thanks. I was wondering, is it possible to prove the last line without using highest weight calculations (which I was trying to avoid)? $\endgroup$ Oct 9, 2015 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.