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In Wikipedia and MathPlanet an equivalent definition of a symplectic matrix is given:

$$\left( \begin{array}{ccc} A & B \\ C & D \end{array} \right)$$

is symplectic if and only if:

$$A^TD-C^TB=I, A^TC=C^TA, D^TB=B^TD$$

but it seems wrong, since, for example:

$$\left( \begin{array}{ccc} 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right)$$

is symplectic but doesn't satisfy the conditions. Or have I mixed everything up?

EDIT: this is crazy talk. That matrix isn't symplectic! (not for the form defined in Wikipedia or MathPlanet.

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Is what you wrote really a symplectic matrix? What is your definition of a symplectic matrix? –  Willie Wong Dec 19 '10 at 21:27

2 Answers 2

This is another question which highlights the problems with not thinking about things in a coordinate-free manner. Symplectic transformations are defined relative to a symplectic form, and symplectic matrices in turn are defined relative to some "canonical" symplectic form with respect to the standard basis. The problem is that there are at least two reasonable choices for such a "canonical" form (both of which are described at the Wikipedia article), and the resulting symplectic matrices you get from each form are different. So you are probably just using a different one.

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Too true :) Can I delete this ridiculous question? –  simplequestions Dec 19 '10 at 21:34
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Unfortunately I don't think you can delete a question if it has an answer with at least 2 upvotes. But I don't see the harm in leaving it up; this is a mistake I can imagine many students making, so it's valuable to have it documented here. –  Qiaochu Yuan Dec 19 '10 at 21:40
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Well, I think Qiaochu can delete this question. But I agree with Qiaochu: for pedagogical purposes it is a very nice example. For the more usual matrix groups (orthogonal/unitary), the canonical bilinear form is already too rigidly stuck in our heads, so it is nice to have an example where the value of "canonical" is up to debate. –  Willie Wong Dec 19 '10 at 21:45
    
Then doesn't it make sense to add the same kind of comment to Wikipedia, mine and others' pedagogy? –  simplequestions Dec 19 '10 at 21:55
    
@simplequestions: this is already implicit in the second sentence of the Wikipedia article. –  Qiaochu Yuan Dec 19 '10 at 22:00

A $2n \times 2n$ matrix $X$ is called symplectic iff it satisfies $X^t J X = J$, where

$$J = \begin{bmatrix} 0 & I \\-I & 0\end{bmatrix}$$ with $I$ being an identity matrix of rank $n$.

It is easy to check by partitioning $X$ into block form $X=\begin{bmatrix}A & B \\C & D\end{bmatrix}$ that this definition is identical to the one mentioned in both Wikipedia and MathPlanet. So the definition is OK!

But the matrix in your case does not satisfy the equation $X^t JX=J$. I strongly recommend that you check it directly.

Cheers!

Richard

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Hi Richard, welcome to this site! You can use LaTeX pretty much as you're used to, simply by enclosing it in dollar signs (click on edited xx mins ago above my name to see the details of what I did). See you around! –  t.b. May 7 '11 at 16:48
    
Hmm, the convention in here (well, the one I am accustomed to at least) seems to be backwards... –  J. M. May 7 '11 at 16:56

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