From Wikipedia:

Let B be a bilinear form with a trivial radical on the space $V$ over the field $K$ with characteristic different from 2. One can now define a map from $D(V)$, the set of all subspaces of $V$, to itself :

$$\alpha:D(V)\rightarrow D(V) :W\mapsto W^{\perp}.$$

This map is an orthogonal polarity on the projective space $PG(W)$. Conversely, one can prove all orthogonal polarities are induced in this way, and that two symmetric bilinear forms with trivial radical induce the same polarity if and only if they are equal up to scalar multiplication.

Orthogonal polarity, a symmetric bilinear form in algebraic geometry

  1. I understand the projective space of a vector space, but I wonder what the notation $PG(W)$ means?
  2. How is $\alpha$ a symmetric bilinear form, given that its domain $D(V)$ is not a product space?

Thanks and regards!


1 Answer 1


Wikipedia is surprisingly weak on these things. I've started to try to improve that a bit, but I think that will be a long-term project. For now, to answer your questions:

The notation $PG(W)$ seems to be used sporadically on Wikipedia without being explained anywhere. I believe it refers to the projective geometry of $W$, which is basically the same thing as $D(V)$, but with the structure of a geometry, i.e. equipped with a symmetric incidence relation. You can find definitions here and here.

That "definition" of orthogonal polarity isn't quite right. A polarity is a map on D(V) that maps lines to hyperplanes and hyperplanes to lines such that its square is the identity. One can show that polarities are in one-to-one correspondence (up to scale factors) with non-degenerate reflexive (or orthosymmetric) sesquilinear forms, where "reflexive" means that the orthogonality relations defined by $s(u,v)=0$ and $s(v,u)=0$ are the same and "non-degenerate" means that no vector is orthogonal to all vectors. Such a sesquilinear form thus induces a symmetric notion of orthogonality, and thereby induces a polarity which maps every vector to the hyperplane orthogonal to it and vice versa. One can further show that a non-degenerate reflexive sesquilinear form must be either an alternating form or a scalar multiple of a hermitian form. This is the Birkhoff-von Neumann theorem (not to be confused with this other Birkhoff-von Neumann theorem); the most lucid exposition and proof of it that I've found on the net is here. (Don't mind the fact that those notes deal only with finite fields; I believe the proof goes through for arbitrary fields.) Polarities are classified according to the sesquilinear forms that induce them. A polarity induced by an alternating form is called symplectic, and a polarity induced by a hermitian form is called hermitian. In the case where the field automorphism associated with the sesquilinear form is the identity, a hermitian sesquilinear form is a symmetric bilinear form, and the corresponding polarity is called an orthogonal polarity.

Thus, it's not that an orthogonal polarity is a symmetric bilinear form, it is induced by one; so your two quotes from the two Wikipedia article are actually trying to say the same thing. (The first one was missing "symmetric" in front of "bilinear", but I've added that since.)


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