Timeline for Do skew symmetric matrices ever naturally represent linear transformations?
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| Jul 24, 2013 at 15:39 | comment | added | Liviu Nicolaescu | @ Frank The infinitesimal action on $\mathbb{R}^n$ of a matrix in the Lie algebra of $O(n)$ is the usual action of a $n\times n$ matrix on $\mathbb{R}^n$. The Lie algebra of $O(3)$ is $3$-dimensional and can be canonically identified with $\mathbb{R}^3$ using the Pauli matrices. Under this identification, the Lie algebra action becomes the cross product. Also, I think that Robert Bryant makes a very good point. | |
| Jul 24, 2013 at 14:27 | comment | added | Frank Thorne | @Robert: Thank you very much. I guess I mean (roughly) an action that has some nice geometric interpretation, or which is studied in some other context. If I understood correctly, your last sentence is not answering my question in the affirmative? | |
| Jul 24, 2013 at 14:24 | comment | added | Frank Thorne | Thanks to all: @Liviu: Guiseppe's example is very nice, but I understand that the cross product exists only in 1, 3, and 7 dimensions? Do you just mean in an infinitesimal sense, or would one ever "naturally" write down the action of a matrix in the Lie algebra of $O(n)$ on $\mathbb{R}^n$? If the latter, I would be very grateful if you would explain further. | |
| Jul 22, 2013 at 17:09 | comment | added | Robert Bryant | I think that the essential question is what you mean by 'naturally'. For a general vector space $V$ over a field, there is nothing natural about skew-symmetric matrices, as this condition is not preserved by an arbitrary choice of basis (which is how one gets a 'natural' correspondence between linear transformations and matrices in the first place). However, for a vector space $V$ endowed with a non-degenerate symmetric inner product that can be written as a sum of squares, the choice of an orthonormal basis for the inner product produces the natural correspondences you mention. | |
| Jul 22, 2013 at 16:35 | comment | added | Liviu Nicolaescu | The matrices in the Lie algebra of $O(n)$ act naturally on $\mathbb{R}^n$. When $n=3$ one obtains the situation described by Giuseppe Negro. | |
| Jul 22, 2013 at 16:26 | comment | added | Giuseppe Negro | In fluid mechanics, the anti-symmetric part of the strain tensor does play a role. And it arises as a transformation of the form $$\mathbf{x}\mapsto \omega \times \mathbf{x},$$ where $\times$ is the vector product of $3$-dimensional space. Does it count as an example? | |
| Jul 22, 2013 at 16:22 | history | asked | Frank Thorne | CC BY-SA 3.0 |