MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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 symplectic space over a field $k$ (for simplicity, the characteristic of $k$ is not $2$). The Maslov index sends a collection of $n$ lagrangian subspaces of $V$ to a quadratic space over $k$ (see Teruji Thomas's excellent paper on this).

My question: is this index surjective on (isometry classes of nondegenerate) quadratic spaces as we allow $n$ and the choice of lagrangians to vary, or if not then which quadratic spaces it hits? I don't think it matters, but I'm interested in the case where $k$ is a non-Archimedean local field.

share|cite|improve this question
For $k=\mathbb{R}$--where the Witt ring has only two elements--I don't even think surjectivity is here so easy to answer, ie. are all Maslov indices hyperbolic? This is a funny paper, borderline between linear symplectic geometry (which I take as a geometry over $\mathbb{R}$) and the theory of quadratic forms which is very much field-theoretic (and almost totally trivial over $\mathbb{R}$). In what context does a symplectic space over, say $\mathbb{R}(x)$ or $\mathbb{C}(x)$? – J. Martel Jun 17 '12 at 6:14
@J. Martel: I'm very confused by your comment. The Witt ring of $\mathbb{R}$ is isomorphic to $\mathbb{Z}$. Certainly it has more than two elements: the two one-dimensional forms $x \mapsto x^2$ and $x \mapsto -x^2$ are non-isometric and nontrivial in the Witt ring. – Justin Campbell Jun 17 '12 at 18:04
Also, this paper is mostly concerned with the case that $k$ is a non-Archimedean local field, e.g. $\mathbb{Q}_p$. – Justin Campbell Jun 17 '12 at 18:14
What I wrote is nonsense. Of course $\lbrace 1 \rbrace, \lbrace -1 \rbrace$ are non-isometric over $\mathbb{R}$. But that's where the story of quadratic forms over $\mathbb{R}$ stops. ie. the fundamental class is cyclic on $\lbrace 1, -1 \rbrace$. The $\mathbb{Z}/2$ i had in mind was the Brauer group of $\mathbb{R}$. On second point, I am without any instance of a linear symplectic space over $\mathbb{C}(t)$ which has symplectic form not just $\omega_{\mathbb{R}} \otimes \mathbb{C}(t)$. – J. Martel Jun 17 '12 at 19:16
An artificial example might arise from the tangent bundle of a symmetric product $Sym^kV$, where $V$ is a projective variety over $\mathbb{C}(t)$. ie. take a tangent space. – J. Martel Jun 17 '12 at 19:18

Your Answer


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

Browse other questions tagged or ask your own question.