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Is there any smooth function that transform trace operator to determinant? More precisely is there any smooth function $f:\mathbf{Mat}_{n}(\Bbb R)\to \mathbf{Mat}_{n}(\Bbb R)$ such that $\det\circ f=\mathscr{tr}$ (or equivalently $\mathscr{tr}\circ f=\det$)?

P.S. I am asking this question because I want to find a new geometric interpretation for trace (other than those mentioned in MO post) using determinant operator.

C.F.G
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    At least, $\det(e^A)=e^{{\rm tr}(A)}$, see here. – Dietrich Burde Nov 21 '22 at 17:20
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    It's not very interesting, but the map $f$ that sends $A$ to the $n \times n$ diagonal matrix with diagonal $(\operatorname{tr} A, 1, \ldots, 1)$ satisfies $(\det \circ f)(A) = (\operatorname{tr} A)(1)^{n - 1} = \operatorname{tr} A$. – Travis Willse Nov 21 '22 at 17:27
  • @TravisWillse, I want to find a new (other than mentioned in MO post) geometric interpretation for trace using det. Your comment is interesting but can it help? – C.F.G Nov 21 '22 at 17:34
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    @C.F.G I don't see how. You should mention that context in the question statement, since it may help you get better answers. It may be helpful that, more generally, for any right inverse $g$ of $\det$, we have $\det\circ (g \circ \operatorname{tr}) = \operatorname{tr}$, i.e., $f := g \circ \operatorname{tr}$ satisfies your condition. – Travis Willse Nov 21 '22 at 17:44
  • @C.F.G Notably, Dietrich's comment gives a smooth function $f$ such that $\det \circ f = f \circ \operatorname{tr}$ – Ben Grossmann Nov 21 '22 at 17:58
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    @BenGrossmann It might be worth emphasizing for O.P. that the two functions $e^{\cdot}$ in the identity Dietrich mentioned are different (at least for $n \neq 1$): One is the scalar exponential, and the other the matrix exponential. – Travis Willse Nov 21 '22 at 21:57

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Disclaimer: The following is not a solution, and may be completely useless. I just followed the assumptions and tried to see what $f$ would need to look like. So might as well post it here as a summary.

So let's say that first, we restrict our attention to $GL_n(\mathbb{R})$, the space of invertible matrices (and then we'll generalize). Then, we have two fiber bundles:

enter image description here

As Dietrich Burde mentioned, the most natural way to fill up this diagram is the following: enter image description here.

However, I think you're looking for a map $f : \text{Mat}_n(\mathbb{R}) \to \text{Mat}_n(\mathbb{R})$ such that $\det \circ f = \text{tr}$. So let's see what that would look like.

  • First, let's make a bad assumption about this map $f$: let's assume it maps invertible matrices to invertible matrices.

In other words, it restricts to $f : GL_n(\mathbb{R}) \to GL_n(\mathbb{R})$ such that $\det \circ f = \text{tr}$. So it gives us a diagram that looks like this:

enter image description here

where the bottom map is the natural inclusion.

However, there are some matrices in the top right version of $GL_n(\mathbb{R})$ that have trace $0$, so whatever the map $f$ is, there is no way you can get this diagram to commute. Contradiction.

Conclusion: if such an $f$ exists, it maps some invertible matrices to non-invertible matrices. (And actually, it's easy to check that it's exactly the matrices with trace $0$ that are gonna be sent to non-invertible matrices. It actually pretty much follows from the definition of $f$).

Now going back to the general $f$. Your map $f : \text{Mat}_n(\mathbb{R}) \to \text{Mat}_n(\mathbb{R})$ would need to map matrices of trace $0$ (which is a closed, negligible set in $\text{Mat}_n(\mathbb{R})$) into the set of matrices of determinant $0$ (which is also a closed, negligible set in $\text{Mat}_n(\mathbb{R})$). And it needs to do so in a smooth way.

I've got to admit that I'm struggling to visualize what that map would look like, because both of the sets are kind of "sparse" in Mat$_n(\mathbb{R})$. It may definitely be possible (or maybe there's something forbidding it), but I can't see anything. Maybe someone will have an idea how to continue from there! (Or maybe this isn't useful at all but it was at least fun to imagine)

Azur
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  • PS: I'm rereading that comment and wondering if it is any useful, or if I just made a long-winded tautological argument. – Azur Nov 21 '22 at 18:46