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Then we can immediately define the "regularized trace" TR A = ζA(-10) and the "regularized determinant" DET A = exp(-ζA'(-10)), where by ζA'(s) I mean the derivative of ζA(s) with respect to s. (If the eigenvalues λn are discrete, then ζA(s) = Σ λn-s, and so one would have TR A = Σ λn and DET A = Σ (log λn) λn-s |s=0, if they converged.) If A is trace- (determinant-) class, then TR A = tr A (DET A = det A).

Then we can immediately define the "regularized trace" TR A = ζA(-1) and the "regularized determinant" DET A = exp(-ζA'(-1)), where by ζA'(s) I mean the derivative of ζA(s) with respect to s. (If the eigenvalues λn are discrete, then ζA(s) = Σ λn-s, and so one would have TR A = Σ λn and DET A = Σ (log λn) λn-s |s=0, if they converged.) If A is trace- (determinant-) class, then TR A = tr A (DET A = det A).

Then we can immediately define the "regularized trace" TR A = ζA(0) and the "regularized determinant" DET A = exp(-ζA'(0)), where by ζA'(s) I mean the derivative of ζA(s) with respect to s. (If the eigenvalues λn are discrete, then ζA(s) = Σ λn-s, and so one would have TR A = Σ λn and DET A = Σ (log λn) λn-s |s=0, if they converged.) If A is trace- (determinant-) class, then TR A = tr A (DET A = det A).

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Theo Johnson-Freyd
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The short answer to may question my question may be a pointer to the right text. I will give all the background I know, and then ask my questions in list form.

The short answer to may question my be a pointer to the right text. I will give all the background I know, and then ask my questions in list form.

The short answer to my question may be a pointer to the right text. I will give all the background I know, and then ask my questions in list form.

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Ilya Nikokoshev
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Theo Johnson-Freyd
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