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I have a couple ideas for the following problem and would like verification, since I am still shaky with representation theory.

Let $V$ be a $n$-dimensional vector space over a field $k$ and let $\Lambda(V)$ be the exterior algebra. Give a non-trivial representation of $GL(V)$ on $\Lambda(V)$. Then fix a basis of $V$ and let $T\subset GL(V)$ be the subgroup identified with the diagonal matrices. Give a decomposition of $\Lambda(V)$ into irreducible subrepresentations.

We obviously have $V$ as a natural representation of $GL(V)$. And thus we have a representation on $T(V)$, the tensor algebra of $V$ (define the action on $k$ to be trivial). A lemma now says that if we have a $I\subset T(V)$ which is $GL(V)$-invariant, we can define a representation on $T(V)/I$ by acting on the representatives. Since $\Lambda(V)=T(V)/I,I=span(v\otimes v),v\in V$ I is clearly invariant so this is gives us a non-trivial representation.

Now I tried to identify the subrepresentations with respect to $T$. If we let $\{v_1,...,v_n\}$ be a basis of $V$, we see that the span of each $v_i$ is $T$-invariant. So the decomposition would be $V=\bigoplus_{i=1}^n span(v_i)$. But doesn't this already give us the decomposition for $\Lambda(V)$, since $\Lambda(V)=\bigoplus_{k=0}^n\Lambda^k(V)$, so we have a finite amount of copies of $V$ "tensorized" together and then modded out a $T$-invariant subspace. Thus we can just distribute the direct sum over the tensor product and are done?

I would really appreciate any advice!

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So the first thing to check is that the action on the exterior algebra is given by $g(v_1\wedge v_2\wedge...\wedge v_i)=g(v_1)\wedge g(v_2)\wedge...\wedge g(v_i)$, this is just because it descends from the action on the tensor algebra. Once you have this it is clear that $\Lambda^k(V)$ is an invariant subspace for each $k$. It turns out that these are all irreducible but that's a bit harder to see.

Now if $e_1, e_2, ... e_n$ are basis vectors for $V$ which are common eigenvalues for $T$ then all vectors of the form $e_{i_1} \wedge e_{i_2}...\wedge e_{i_k}$ with $i_1<i_2...<i_k$ will also be common eigenvectors for $T$ in this representation. These span $\Lambda^k(V)$ so we see this gives a decomposition of $\Lambda^k(V)$ as a $T$ representation.

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  • $\begingroup$ Isn't the reasoning behind that the action descending from the action on the tensor algebra exactly that the ideal we mod out is invariant? I also see that the $\Lambda^k(v)$'s are clearly subrepresentations, but I don't see why they are irreducible? $\endgroup$ – blst May 11 '14 at 0:56
  • $\begingroup$ Right, you need to quotient by an invariant space, I was just saying that in this particular case we can be more explicit with how we act is all. To see they are irreducible you want to use the decomposition for $T$. Any submodule will also need to be a $T$ submodule, and will hence have to be a direct sum of the subspaces spanned by the basic wedges $e_{i_1} \wedge e_{i_2}...\wedge e_{i_k}$, then you just need to show that $GL(V)$ acts transitively on these wedges. $\endgroup$ – Nate May 11 '14 at 1:12
  • $\begingroup$ I'm with you up until the last part: Clearly a subrepresentation has to be a $T$-submodule and are spanned by the basic wedges, but why do we need to show that $GL(V)$ acts transitively? (not $T$, where it isn't the case. Note that we are looking for a decomposition as a representation of $T$.) $\endgroup$ – blst May 11 '14 at 1:26
  • $\begingroup$ Perhaps "acts transitively" isn't the right way to phrase it since indeed it doesn't act on the set of such wedges. What I mean is that there is an element of $GL(V)$ that sends any one such wedge to any other, so any invariant subspace that contains one must contain all of them. $\endgroup$ – Nate May 11 '14 at 1:41
  • $\begingroup$ But it doesn't have to be $GL(V)$ invariant, just $T$ invariant, which for example $\Lambda^0(V)=V$ isn't. This is why I'm confused as to why the $\Lambda^k(V)$ give a decomposition into irreducible subrepresentations. $\endgroup$ – blst May 11 '14 at 1:41

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