Let $V$ be a vector space over an algebraically closed field, equipped with a symmetric, non-degenerate, bilinear form $\beta : V \times V \to \mathbb{C}$. Let $O \subset GL(V)$ be the group of linear transformations preserving $\beta$. $O$ is the orthogonal group.
Is $V$ irreducible as a representation of $O$?
I can prove this using a fact from algebraic geometry, but this only works over $\mathbb{C}$ (as far a I know).
Proposition: Work over $\mathbb{C}$. Let $X$ be a the quadric hypersurface of $\{v : \beta(v,v) = 0\}$. Then $X$ is a homogeneous space for the action of $O$, that is, the action of $O$ is transitive.
Pf:
We can assume that $\beta(v,w) = \Sigma v_i w_i$. Let $G^+$ be the space of oriented $2$ planes in $\mathbb{R}^n$, equipped with the usual dot product. For an orthonormal basis $v_1, v_2$, the point $w = v_1 + i v_2 \in \mathbb{C}^n$ satisfies $w^T w = 0$. A different choice of orthonormal basis would changed $w$ by a phase, and $w$ determines the orthonormal basis up to phase. Morover, if $w = v_1 + i v_2$ satisfies $w^T w = 0$, it is easy to see that $\alpha v_1, \alpha v_2$ are orthonormal, for some $\alpha$. Thus there is a correspondence between $G^+$ and the (projective) quadric hypersurface $X$.
This correspondence shows us how to prove that the action is transitive. Let $[x],[y] \in X$, with $x = x_1 + x_2 i$ and $y = y_1 + y_2 i$ so that $x_1, x_2$ and $y_1, y_2$ are each an orthonormal pair of vectors in $\mathbb{R}^n$. Then there is an $A \in O(\mathbb{R})$ so that $Ax_1 = y_1$ and $Ax_2 = y_2$. This $A$ lifts to an action on $\mathbb{C}^n$ by $A(v_1 + i v_2) = Av_1 + i Av_2$. Moreover, this action is in $O(\mathbb{C})$, which can be seen by explicitely comparing $\beta( A(s + it) , A(x + iy))$ with $\beta(s + it, x + iy)$. QED
Lemma: The action of $O$ is transitive on $X^c$. (This is true because any $v \in V$ with $\beta(v,v) \not = 0$ can be extended into an orthonormal basis for $\beta$. "QED".)
So if $W \subset V$ is a subrepresentation, and $W \not = 0$, then one of $\mathbb{P}(W) \cap X$ or $\mathbb{P} \cap X^c$ is non-empty. Transitivity of the $O$ action and invariance of $W$ now implies that $\mathbb{P}(W) \supset X$ or $\mathbb{P}(W) \supset X^c$, and in both cases this implies that $W = V$.
Thus $V$ is irreducible as an $O$ representation, at least over $\mathbb{C}$.
To reiterate my questions:
- Is there an "easier" way to show that the defining representation of $O$ is irreducible? A more algebraic argument? (The isomorphism between $X$ and the Grassmannian of oriented planes is not obvious without hindsight, and seems to be only true over $\mathbb{C}$, or fields with some similar properties.) Not that I dislike geometry - I prefer it - but I wouldn't be surprised if there was an easier manpulation I was missing.
- Is it still true over other fields? Algebraically closed fields of positive characteristic? Finite fields?