# Exercise at the Beginning of Part II in Fulton's Book on Young Tableaux

In Fulton's Book Young Tableaux, there's an Exercise at the beginning of part II for which I cannot find a solution (there doesn't seem to be one for this exercise in my copy of the book). It reads:

Exercise. Show that, if $e_1,\ldots,e_m$ is a basis for the ($\mathbb{C}$-vectorspace) $E$, then the images of the vectors $(e_i\wedge e_j)\otimes e_k$, for all $i<j$ and $i\le k$, form a basis of $E^{(2,1)} := \left.{\textstyle\bigwedge^2E}\otimes E\middle/\left((u\wedge v)\otimes w - (w\wedge v)\otimes u - (u\wedge w)\otimes v\:\middle|\:u,v,w\in E\right)\right..$

First, I do not see why, for $i<j<k$, the equality $(e_i\wedge e_j)\otimes e_k = (e_k\wedge e_j)\otimes e_i + (e_i\wedge e_k)\otimes e_j$ already holds in $\bigwedge^2 E\otimes E$. But even that aside, I do not see how to solve the exercise. Any help would be welcome.

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The equality $(e_i\wedge e_j)\otimes e_k = (e_k\wedge e_j)\otimes e_i + (e_i\wedge e_k)\otimes e_j$ definitely does NOT already hold in $\wedge^2 E \otimes E$. –  darij grinberg Apr 6 '12 at 22:16

It is enough to show two things:

1) The listed element generate the space.

We know that $(e_i \wedge e_j) \otimes e_k$ with no restraints on the indices generate the space, and we can clearly restrict to $i < j$ by the antisymmetric behaviour of the wedge product. Now, if $i > k$, we use the identity:

$$((u∧v)\otimes w=(w∧v)⊗u+(u∧w)⊗v$$

for $u=e_i$, $v=e_j$ and $w=e_k$.

This gives $$((e_i∧e_j)\otimes e_k=(e_k∧e_j)⊗e_i+(e_i∧e_k)⊗e_j= (e_k∧e_j)⊗e_i-(e_k∧e_i)⊗e_j,$$

which shows that the vector is actually a linear combination of two of our given vectors. Therefore, they are indeed a generating set.

2) The listed elements are independent.

Note that the relations in the ideal are multilinear in $u$,$v$,$w$, so it is enough to use the relations for the basis vectors to generate the full ideal.

Now, suppose that you have a linear combination of the given vectors that gives zero in the quotient space. This means that the linear combination lies in the ideal which means that the linear combination is a linear combination of relations in the ideal.

You can now look at the smallest index of an $e_i$ that occurs in this expression and check that its coefficient is 0.

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You should know that the $e_i \wedge e_j (i<j)$ are a basis of $\bigwedge^2E$, so that after tensoring you have a basis $$e_i \wedge e_j \otimes e_k \quad(1\leq i< j \leq n,1\leq k \leq n)$$ for $\bigwedge^2E \otimes E$.

All you have to show is that the elements where $k<i$ are in the subspace $V$ generated by $$(u\wedge v) \otimes w - (w\wedge v)\otimes u - (u\wedge w)\otimes v\$$ but look at

$$(e_i \wedge e_j ) \otimes e_k \equiv (e_k \wedge e_j)\otimes e_i + (e_i\wedge e_k)\otimes e_j \quad (\text{mod } V)$$

and other permutations of the indices. Assume $k<i<j$. Can you show that the left side is $\equiv 0$ mod $V$?

EDIT: this is not an answer, the necessary computations are more difficult than I expected!!

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It is not sufficient to show what you just said. We have to show that $e_i\wedge e_j\otimes e_k$ is in the subspace $V$ if and only if $k<i$. That is what my first question was about, and then you simply rephrased my second question, without really answering it. –  Jesko Hüttenhain Apr 6 '12 at 21:21
You are right, I forgot the only if part. But I did not rephrase your second question, because you only have to show this mod $V$! –  Blah Apr 7 '12 at 12:22
Well, your answer still isn't an answer, but a question. I will answer it: I can not show that the left or the right side are $0$ modulo $V$. If you can, that'd be at least half an answer. –  Jesko Hüttenhain Apr 8 '12 at 6:21
Another question: can you see from the above formula that the elements with $i\leq k$ generate the quotient space because you can always move the third index? –  Blah Apr 8 '12 at 8:09