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In this paper(http://arxiv.org/pdf/1204.6131.pdf), the following statement

$V =K^n$ and its dual are isomorphic $SL(2,K)$-modules,

seems to be common sense. Here, $K$ is a field of characteristic $0$.

I don't know if $G = SL(2,K)$ acts on the dual of $V$, namely $V^*$, in this fashion:

$(gf)(x) = f(g^{-1}x)$ for any $g \in G$, $f \in V^*$ and $x \in V$.

If this is the case, let $e_1, \cdots, e_n$ be a basis of $V$, and $x_1, \cdots, x_n$ be a dual basis of $V^*$, that is, $x_i(e_j) = \delta_{ij}$, then for any $g \in G$, $g(e_i) = \sum_{j=1}^n a_{ji}e_j$ for a matrix $A = (a_{ij}) \in GL(n,K)$. Let $A^{-1} = (a_{ij}')$ be its inverse.

Suppose that $V$ and $V^*$ are isomorphic $G$-modules and $\phi: V \rightarrow V^*, e_i \mapsto \sum_{j=1}^n b_{ji}x_j$ gives such an isomorphism. Then as $(gx_i) (e_j) = x_i(g^{-1}e_j) = x_i(\sum_{k=1}^n a_{kj}'e_k) = a_{ij}'$, $g(x_i) = \sum_{j=1}^n a_{ij}'x_j$.

So $g(\phi(e_i)) = g(\sum_{j=1}^n b_{ji}x_j) = \sum_{j=1}^n b_{ji}(gx_j) = \sum_{j=1}^n b_{ji} (\sum_{k=1}^n a_{jk}' x_k) = \sum_{j,k=1}^n b_{ji} a_{jk}' x_k$, and $\phi(g(e_i)) = \phi(\sum_{j=1}^n a_{ji} e_j) = \sum_{j=1}^n a_{ji} \phi(e_j) = \sum_{j=1}^n a_{ji} (\sum_{k=1}^n b_{kj} x_k) = \sum_{j,k=1}^n a_{ji} b_{kj} x_k$.

They are equal, so if I let $B = (b_{ij})$, then $B^T A^{-1} = A^T B^T$, that is $BA = (A^{-1})^T B$.

But if I let $K = \mathbb C$, and $G$ acts on $\mathbb C^2$ in the natural way, it seems impossible to find $B \in GL(2, \mathbb C)$ such that $BA = (A^{-1})^T B$ for any $A \in G$. (For example, there is no such common $B$ for $A_1 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $A_2 = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$.)

So, the statement in the paper does not seem to be exactly the case. But I trust the author and I believe there must be something wrong in my method or calculation. Would anyone please tell me where I am wrong?

Moreover, the author explained to be in the language of the representation theory of Lie algebras. I would be appreciate if someone is willing to explain to me the correspondence of representation of the algebraic group and the Lie algebra in case of $SL(2,K)$.

Thanks a lot.

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    There is something wrong with my internet connection, so I can't see the page normally. So please forgive me if I have made any mistake in Tex.2012-06-19
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    It *is* possible to find a $B$ with the properties you want for the $n=2$ case: $B=\left(\begin{array}{cc} 0&1 \\ -1&0 \end{array}\right)$. For higher $n$, though, this might become much harder. The standard constructive proof for arbitrary $n$ goes by constructing an $\mathrm{SL}_2$-invariant bilinear form on $V$; how exactly this is done depends on how you define the $\mathrm{SL}_2$-module $V$. Alternatively, you can probably use the representation theory of $\mathrm{SL}_2$ to show that there exists only one irreducible $\mathrm{SL}_2$-representation of any given dimension; but I don't ...2012-06-19
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    ... know the details (it might very well be harder than for the corresponding Lie algebra).2012-06-19
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    If you define $V$ as the module of binary forms of degree $n$, with $\mathrm{SL}_2$ acting by coordinate transformations, then an $\mathrm{SL}_2$-invariant bilinear form $B$ is given by $B\left(\sum\limits_{i=0}^n a_i x^i y^{n-i},\sum\limits_{i=0}^n b_i x^i y^{n-i}\right) = \sum\limits_{k=0}^n \left(-1\right)^k k! \left(n-k\right)! a_k b_k$. Or, more abstractly: $B\left(P,Q\right)$ is the coefficient before $x^0y^0$ of the polynomial $\left(P\left(D_y,-D_x\right)\right)Q$, where $D_x$ and $D_y$ are the derivations with respect to $x$ and $y$, and $P\left(D_y,-D_x\right)$ means "the ...2012-06-19
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    ... polynomial $P$ applied to the (commuting!) operators $D_y$ and $-D_x$". The equivalence of these two definitions of $B$ is easy to check, and $\mathrm{SL}_2$-invariance is nicer to prove on the second definition (but more straightforward on the first, if you know how to work with binomial coefficient sums).2012-06-19
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    Oops, I just see that my $V$ was $K^{n+1}$ instead of $K^n$. But I assume that you won't have any difficulties with renaming my variables.2012-06-19
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    Also, let me take the liberty to remind people about a vaguely related question of my own ( http://mathoverflow.net/questions/38165/why-is-every-polynomial-representation-of-sl2-selfdual ) about how to prove selfduality for *any* polynomial representation of $\mathrm{SL}_2$ without having to break it up into irreducibles. I'd still like to know the answer.2012-06-19
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    It's important to note that this is false in positive characteristic, so that puts a lower bound on how trivial a proof of this result can be (it can't be the result of abstract tensor manipulations, for example, that would be valid over any field).2012-06-19
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    Thanks to @darij grinberg for taking time to write the comments, which give me new insight into this problem.2012-06-24

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