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I know that every (Lie group) representation of $\textrm{GL}_n(\Bbb{C})$ is completely reducible; this I believe comes from the fact that every representation of the maximal compact subgroup $\textrm{U}(n)$ is completely reducible. More explicitly, suppose $V$ is a representation of $\textrm{GL}_n(\Bbb{C})$. Then $V$ is also a representation of $\textrm{U}(n)$, by complete reducibility of the unitary group we know that there is a $\textrm{U}(n)$ invariant inner product such that if $U$ is any $\textrm{GL}_n$ - invariant subspace of $V$ (and hence $\textrm{U}(n)$ invariant), there is an orthogonal complement $W$ such that

$$V = U \oplus W$$

with $W$ invariant under $\textrm{U}(n)$. Now $W$ as a representation of the real Lie algebra $\mathfrak{u}(n)$ is invariant and hence under the complexified Lie algebra $$\mathfrak{gl}_n = \mathfrak{u}_n \oplus i \hspace{1mm} \mathfrak{u}(n).$$

Since $\textrm{GL}_n(\Bbb{C})$ is connected $W$ is also invariant under $\textrm{GL}_n$ showing that every representation of it is completely reducible.

Now I have read several textbooks on representation theory (e.g. Bump's Lie Groups, Procesi's book of the same name) and they all seem to tacitly assume that every representation of $\textrm{GL}_n$ is completely determined by its character; i.e. if two representations have the same character then they are isomorphic.

Now in the finite groups case, we concluded this fact based on 1) Maschke's Theorem and 2) Linear independence of characters.

We do not necessarily have 2) so how can we conclude the fact I said about about $\textrm{GL}_n$?

Thanks.

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3 Answers

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The character of a representation $V$ tells you how a maximal torus $T$ acts on $V$. Moreover, you can then figure out how any maximal torus acts on $V$ because any other maximal torus is of the form $g T g^{-1}$ for $g \in G$. (Here, $G = GL_n(\mathbb{C})$.) That is, if $v_1, \dots, v_n$ are the $T$-eigenvectors of $V$, then $g \cdot v_1, \dots, g \cdot v_n$ are the $gTg^{-1}$-eigenvectors of $V$.

In the case of $GL_n(\mathbb{C})$, this says that the character determines how every diagonalizable element $g \in GL_n(\mathbb{C})$ acts on $V$. But the diagonalizable matrices are dense in $GL_n(\mathbb{C})$, and thus representations are uniquely determined by their characters.

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Thanks for your answer, I have learned something new! Where can I go to read more about this wonderful facts? –  fpqc Oct 26 '12 at 23:42
    
Why is it the case that the character determines how a maximal torus acts on $V$? Also, I am beginning to wonder if the answer to my original question boils down to the fact that it is true for $\textrm{U}(n)$ and hence for $\textrm{GL}_n$..... –  fpqc Oct 27 '12 at 0:56
    
A Schur polynomial is a non-negative sum of monomials. Each monomial corresponds to an eigenvalue and the multiplicity that it occurs is the coefficient of the monomial. A good source for all of this is Fulton's Young Tableau. –  Michael Joyce Oct 27 '12 at 1:52
    
I am reading the relevant section in Fulton and Harris. I will get back to you as soon as possible. –  fpqc Oct 27 '12 at 12:15
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Ah yes I think I see it now! Suppose we know that every representation of $\textrm{GL}_n$ decomposes into a direct sum of its weight spaces. Then we will know that the action of the maximal torus on the representation is diagonal, so in particular once we know the sum of the eigenvalues we know how the maximal torus acts on the representation. –  fpqc Oct 28 '12 at 22:36
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This boils down to facts about the representation theory of compact groups: there every complex representation is determined by its character. Now a representation of $GL(n,\mathbb C)$ is determines by a representation of its Lie algebra. But this is the complexification of $u(n)$ and complex representations of a Lie algebra are in one to one correspondence with complex representations of its complexification.

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Thanks for your answer. Now suppose we have $\phi,\rho$ two reps of $GL_n$ of equal character. Then their characters when restricted to the subgroup $U(n)$ are also equal and hence are isomorphic as $\textrm{U}(n)$ representations. Why does it follow that they are isomorphic as $\textrm{GL}_n$ representations as well? –  fpqc Oct 28 '12 at 2:44
    
@BenjaLim: Look at the Lie algebra level. A complex representation of $gl(n) \simeq u(n) \otimes \mathbb C$ is determined by how it acts on the real form $u(n)$ –  Eric O. Korman Oct 28 '12 at 3:08
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One does have linear independence of characters. The characters of the various irreps. form a basis for the space of conjugation invariant algebraic functions on $GL_n(\mathbb C)$.

Added some time later: The ring of conjugation invariant functions on $GL_n(\mathbb C)$ is precisely the ring of symmetric functions in the eigenvalues (with the inverse of the determinant, i.e. the inverse of the product of all the eigenavlues, adjoined), which is generated as a polynomial ring by the elementary symmetric polynomials in the eigenvalues (with the inverse of the product $\lambda_1\cdots\lambda_n$ adjointed).
Thus this ring is spanned as a $\mathbb C$-vector space by the monomials in the elementary symmetric polynomials.

The Grothendieck semiring of isomorphisms classes of finite dim'l reps. of $GL_n(\mathbb C)$ is generated by the fundamental representations $\wedge^i ($standard rep$)$, for $i = 1,\ldots,n$, together with the inverse of the last of these (the $n$th exterior power of the standard rep. is just $\det$, and so I am just saying that we throw in $\det^{-1}$ as well as $\det$).

The character of $\wedge^i$ of the standard rep. is the $i$th symmetric polynomial in the eigenvalues, and now by inductively writing "monomials" in the $\wedge^i$ (i.e. tensor products of the various $\wedge^i$) in terms of irreps., it is easy to verify my claim that the characters of the irreps. span the conjugation invariant functions.

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Thanks for your answer. Where can I find information about linear independence of characters of Lie group representations of $\textrm{GL}_n(\Bbb{C})$? Thanks. –  fpqc Oct 27 '12 at 1:11
    
@BenjaLim: Dear Benjamin, I added a sketch of a proof. Regards, –  Matt E Nov 27 '12 at 3:09
    
This can be found in Fulton Harris. –  Bombyx mori Nov 27 '12 at 3:14
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