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Let $V$ be a finite dimentional vector space over a field $k$. Let $g(\cdot,\cdot)$ be a nondegenerate symmetric bilinear form on $V$. Let $O(V)$ be the subgroup of $GL(V)$ that preserves $g$. Then $V$ can be viewed as a reprentation of $O(V)$. I guess this representation is irreducible. Would anyone please give a proof or a counterexample?

Thanks very much!

I don't know what conditions should be imposed on the underlying field $k$. Maybe you need to assume that $k$ is algebraically closed with characteristic 0.

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2 Answers 2

up vote 1 down vote accepted

It is just a matter of showing that given a non trivial subspace $V' \subset V$, you can find an orthogonal transformation that do not stabilize $V'$.

If $\text{car} \neq 2$ you could use orthogonal reflections.

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:Thanks! Let me fill in the details: If $W$ is a subrepresentation, then choose a vector $v$ which is not in $W$. That reflecting about $v$ stablizes $W$ means that $v$ is orthogonal to $W$. Meanwhile $W$ is isotropic because if $g(v,v)$ is nonzero, we can extend it to an orthonormal basis. This contradicts the nondegeneracy of $g$. Is that right? –  Yang Zhou Jul 26 '12 at 15:27

At least if $k$ has characteristic different from $2$, the natural representation of $O(V)$ on $\operatorname{GL}(V)$ is always irreducible.

Step 1: It suffices to assume $k$ is algebraically closed.

Proof: Indeed, although base extension may make an irreducible representation become reducible, if $W \subset_k V$ is a nonzero, proper $O(V)$-invariant subspace, then $W_{\overline{k}}$ is a nonzero, proper $O(V)_{\overline{k}}$-invariant subspace.

Step 2: By Theorem 25 Proposition 30 from these notes, $O(V)$ acts transitively on the set of one-dimensional isotropic subspaces of $V$ and also (since every element of $k^{\times}$ is a square) on the set of one-dimensional anisotropic subspaces of $V$. It follows that the only possible invariant subspaces are $W_1$, the span of all the isotropic vectors, and $W_2$, the span of all the anisotropic vectors. Just by diagonalizing the form, we see that by nondegeneracy $W_2 = V$. If $\operatorname{dim} V = 1$ then there are no isotropic vectors and $W_1 = \{0\}$. If $\operatorname{dim} V$ is even, then $V$ is a direct sum of hyperbolic planes and thus $W_1 = V$. If $\dim V$ is odd and greater than one, choose two linearly independent anisotropic vectors $v_1$ and $v_2$. Then $v_1^{\perp}$ and $v_2^{\perp}$ are distinct codimension one subspaces which are both hyperbolic and thus spanned by isotropic vectors, so $W_1$ contains two distinct hyperplanes and is therefore all of $V$.

Note that conversely, if the bilinear form is identically zero then the orthogonal group is $\operatorname{GL}(V)$ and thus the representation is certain irreducible. However, if the bilinear form is degenerate but not identically zero, then $V^{\perp}$ is a proper, nontrivial invariant subspace. Note also that there need not be any other invariant subspace, i.e., the representation need not be semisimple / completely reducible. The two-dimensional quadratic space with $q(x,y) = y^2$ gives an example of this.

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Thanks very much for your detailed answer and reference. I have been confused by some details of the action of $O(V)$ for a long time! –  Yang Zhou Jul 26 '12 at 15:29

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