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For a self map $f$ on a topological space $X$ we replace "trace" with "determinant" in the alternative Lefschetz formula $$\Lambda(f)=\sum(-1)^i trace(f^*)|H^i(X,\mathbb{Q})$$

So we have $$\Lambda'(f)=\sum(-1)^i Det(f^*)|H^i(X,\mathbb{Q})$$

What kind of dynamical information we can get from this invariant?(This invariant or any other invariant by replacing trace with some other invariant polynomials,i.e. the coefficients of characteristic polynomials)

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    $\begingroup$ What's the motivation for taking the sum of the determinants? If anything I would expect to take some alternating product of them. $\endgroup$
    – Denis Nardin
    Commented Jun 9, 2019 at 20:44
  • $\begingroup$ @DenisNardin yes yes exactly. To be honnest I was thinking to this product about 2 hours ago. But i confess I did not have a very precise motivation for that. My motivation was the following: I was asking myself"Why is trace a natural object in Lefschetz formula?why there is no other formula?" Any way the product suggestion of you is intersting. BTW my question on naturality of trace in Lefchets formula imply that I should review the detail of the Lefschets fixed point theorem. Any way it would be interesting to find some new criterion in fixed theory not involving trace.what will happen $\endgroup$
    – Ali Taghavi
    Commented Jun 9, 2019 at 21:08
  • $\begingroup$ If we consider product? $\endgroup$
    – Ali Taghavi
    Commented Jun 9, 2019 at 21:09
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    $\begingroup$ If all you want is a motivation for the Lefschetz trace formula, then $\Lambda(f)$ has indeed a very natural interpretation: it is the trace of $f^*:C^*(X)→C^*(X)$ in the derived category (there is a definition of trace of an endomorphism of a dualizable object in the derived category, it's formally the same as the definition of trace: if $f:X→X$, the trace is the composite of $1→X⊗X^\vee→X⊗X^\vee→1$, where the middle map is $f⊗1_{X^\vee}$). Moreover $\Lambda(f)$ counts the fixed points with multiplicity, and I don't think you can do better than that just from the homotopy type. $\endgroup$
    – Denis Nardin
    Commented Jun 9, 2019 at 22:03
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    $\begingroup$ Two remarks: one, you already noted, that why not then consider $\text{tr}(\wedge^k)$, which for the extreme cases yields both trace and determinant; and two, integrate a suitable trace to get a determinant. Conversely, by studying some suitable "derivative" of the form with the determinant (as you propose), does one recover the original trace formula (by using a suitably modified version based on determinants)...anyhow, maybe all of what I said is just junk! $\endgroup$
    – Suvrit
    Commented Jun 9, 2019 at 22:16

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If the self map f is the identity, this is just the index which is a "supertrace" as is seen from this formula. Reidemeister torsion/ Analytic torsion would be the "superdeterminant" version, which corresponds to taking an alternating product (with some powers added in). If you want information on the self map f, you can take the torsion invariant of the mapping cylinder.

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  • $\begingroup$ If $f$ is the identity $\det (f^*)$ is 1, so $\Lambda'(f)$ is 0 or 1 according to the parity of $\dim(X)$. $\endgroup$
    – abx
    Commented Oct 19, 2020 at 15:11
  • $\begingroup$ So the torsion invariants are defined using alternating products as people have discussed in the comments in the question. For the Reidemeister torsion, it's easier to see this in the case where the complex is acyclic. $\endgroup$
    – Gibbon
    Commented Oct 19, 2020 at 18:22
  • $\begingroup$ For instance, see definition 4 in these notes. maths.ed.ac.uk/~v1ranick/papers/torsion.pdf In analytic torsion, you form an alternating product after taking some powers of the regularized determinant. en.wikipedia.org/wiki/Analytic_torsion The Cheeger Muller theorem relates these two invariants under certain conditions. $\endgroup$
    – Gibbon
    Commented Oct 19, 2020 at 18:28
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    $\begingroup$ If you're interested in relations to dynamical systems, check out the Fried conjecture. $\endgroup$
    – Gibbon
    Commented Oct 19, 2020 at 19:02
  • $\begingroup$ Thank you for your answer and comments. I just read your answer, I did not receive a nitification but I find it incidently $\endgroup$
    – Ali Taghavi
    Commented Oct 23, 2020 at 22:16

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