Suppose we have a (non-trivial) representation of some special orthogonal group $SO(p,q)$ over a real vector space $V$, I.e. the action of elements of $SO$ leave invariant a non-degenerate symmetric bi-linear form $G$: $$ G(x,y) = G(Rx, Ry) \quad \forall x,y \in V,\>\> \forall R \in SO $$ or in matrix form $$ \{R^T G R = G \> | \> R \in SO\} $$

Now suppose we perform a unitary change of basis $U$, which forces us to, in general, enter the complexificiation of our vector space (or at least sit skew in some subspace (isomorphic to $V$) of the complexification of $V$)

With this change we can extend the bilinear form to a sesquilinear form: $$ G(a+ib,c+id) = G(a,c)+G(b,d)+iG(a,d)-iG(b,c) $$ Extending our orthogonal structure into a unitary structure (using primes to indicate elements in the new basis):

$$ (R^\prime)^\dagger G^\prime R^\prime = (U^\dagger R\>U)^\dagger (U^\dagger G\>U) (U^\dagger R\>U) = U^\dagger (R^T G R)U = U^\dagger G \>U = G^\prime $$ I.e. the $R^\prime$ in the new basis leave $G^\prime$ invariant as a sesquilinear form (inner product)

We can alternatively extend the form to be complex linear: $$ G(a+ib,c+id) = G(a,c)-G(b,d)+iG(a,d)+iG(b,c) $$ Defining a skew-hermitian form $g = iG$, and $\tilde{g}=\>U^T g \>U$, it can be seen there is also a symplectic$^1$ structure, consider: $$ (R^\prime)^T \tilde{g} R^\prime = (U^\dagger R\>U)^T (U^T g\>U) (U^\dagger R\>U) = U^T(R^T g R)U = U^T g \>U = \tilde{g} $$ (recalling that $U^TU^* = 1$.) I.e. the same set of group elements $R^\prime$ simultaneously respect $\tilde{g}$ as a bi-linear form.

Is there a name for this splitting of orthogonal structure into unitary and symplectic structures? Does it have anything to do with the $2$-out-of-$3$ property? Or perhaps a Khaler metric being inherited from a Riemannian metric?

1: I'm not totally sure this deserves the name symplectic, since I always see symplectic forms defined by skew-symmetry not hermiticity. Is there a better name for this? Perhaps it's (factor of $i$ aside) still an orthogonal structure?

EDIT: My suspicion, based on the answer linked below, is that by "entering the complexification" we gain access to the invariant bi(sesqui)-linear forms there. However this is interesting because if that is true then it seems to imply finding real forms may be something equivalent to finding bases where multiple invariant forms collapse into a single one. See: All invariant forms of (a representation of) a semi-simple lie group

  • $\begingroup$ Sorry, I don't fully follow what you do in latest edits. Beyond that, well there is that basic fact (which you're using) that $G(x,y)$ is a hermitian form if and only if $i\cdot G(x,y)$ is a skew-hermitian form, so in this way it's unsurprising that a group action that's invariant to one is also invariant to one of the other, they sort of carry the same information. $\endgroup$ Commented Feb 28, 2022 at 1:53

1 Answer 1


It is clear to me now. When we have a representation $\rho$ of some group $O$ acting on $V$, and $V$ is equipped with an $O$ invariant bilinear form $G$, when we complexify $V$ we introduce the anti-linear map (call it $\sigma$) of conjugation. This conjugation acts obviously as a real-structure preserving the subspace $V = V_\mathbb{R} \subset V_{\mathbb{C}}$. When we perform a change of basis, we are pushing that subspace around in the larger complexified space $V_{\mathbb{C}}$, and the conjugation map changes with it, updating appropriately.

More importantly for the question I had nearly 2 years ago: now we have a compatible triple! Since $G$ and $\sigma$ are $O$-invariant structures, so too is their composition. Thus, when we complexify a vector space with an invariant bi-liner form, and inherit the natural conjugation from the complexification, we will acquire a Hermitian form via $h(\cdot,\cdot) = B(\sigma(\cdot), \cdot)$!


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