Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
: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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.