Timeline for Categorification of determinant
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23 events
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| Mar 17, 2021 at 18:53 | answer | added | Nacho Darago | timeline score: 8 | |
| Mar 17, 2021 at 16:05 | vote | accept | Nalan | ||
| Mar 17, 2021 at 15:33 | answer | added | David C | timeline score: 16 | |
| Mar 17, 2021 at 14:59 | history | edited | Nalan |
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| Mar 17, 2021 at 13:28 | comment | added | Nalan | Ganter and Kapranov have defined symmetric and exterior powers of k-linear categories. Perhaps using their definitions both the determinant and the characteristic polynomial can be defined in these categories. | |
| Mar 17, 2021 at 12:35 | comment | added | Steve Huntsman | I suspect a more fruitful way of considering this problem is to deal with the characteristic polynomial rather than the determinant per se. Recall (en.wikipedia.org/wiki/Characteristic_polynomial#Properties) that the characteristic polynomial of the matrix $A$ can be written as $p_{A}(t)=\sum _{k=0}^{n}t^{n-k}(-1)^{k}\operatorname {tr} (\wedge ^{k}A)$, where exterior powers are indicated. The traces in this sum can in turn be expressed as determinants, as the Wikipedia link shows. | |
| Mar 17, 2021 at 5:22 | comment | added | მამუკა ჯიბლაძე | @BertramArnold There is a decent version over noncommutative division rings, first constructed by Dieudonné ncatlab.org/nlab/show/Dieudonn%C3%A9+determinant | |
| Mar 17, 2021 at 4:28 | history | edited | Nalan |
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| Mar 17, 2021 at 4:10 | comment | added | Alec Rhea | @NeilStrickland If I remember correctly, after defining the superadjugate of an $(n|n)\times(n|n)$ supermatrix $X$ we do have identities like $sAdj(X)_{ij}=\partial_{ij}Ber(X)$ for $j\leq n$, and $sAdj(X)_{ij}=\partial_{ij}Ber^*(X)$ for $j>n$ where $Ber^*$ denotes a canonically defined parity dual of the Berezinian. Actually that might have been what we were trying to prove, I don't quite remember, but in any event identities like this were not ruled out circa 2016. | |
| Mar 17, 2021 at 2:44 | history | became hot network question | |||
| Mar 17, 2021 at 0:04 | history | edited | Nalan |
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| Mar 16, 2021 at 23:38 | comment | added | Neil Strickland | @AndréHenriques I consider the formal properties of the Berezinian to be mediocre because it is only defined for automorphisms, and has no interpretation in terms of an analogue of exterior algebra. As far as I am aware there is nothing like the characteristic polynomial or the Cayley-Hamilton theorem, and nothing like the formula $A.\text{adj}(A)=\det(A).I$. | |
| Mar 16, 2021 at 23:00 | answer | added | John Rognes | timeline score: 17 | |
| Mar 16, 2021 at 22:53 | comment | added | Bertram Arnold | With regard to the original question, note that nice properties of the determinant require commutativity of the underlying ring, so it is reasonable to expect a categorification to take a symmetric monoidal ($\infty$-)category $C$ as its input. In this case the $K$-theory spectrum acquires a $E_\infty$-ring spectrum structure, and I'd guess that $gl_1(K(C))$ should be the universal recipient of a determinant map; this is at least consistent with the example of super vector spaces and the fact that the higher $K$-groups are recipients of determinant-like invariants like Reisemeister torsion. | |
| Mar 16, 2021 at 22:45 | comment | added | Bertram Arnold | @AlecRhea the signs can usually be hidden efficiently in the symmetric monoidal structure by defining the Berezinian as a functor from super vector spaces and isomorphisms to super lines. Of course, taking values in a (Picard) groupoid makes the Berezinian formally different from the determinant; another big difference is that it does not extend to the whole matrix algebra. | |
| Mar 16, 2021 at 22:26 | comment | added | Theo Johnson-Freyd | @NeilStrickland Emphasizing up on Andre's comment, the superdeterminant aka Berezinian seems to this category theorist to have perfectly good formal properties: it is the universal map of super Lie groups $GL(m|n) \to GL(1)$, etc. What properties do you find mediocre? | |
| Mar 16, 2021 at 22:13 | comment | added | Alec Rhea | @AndréHenriques I worked computationally with Berezinians for some undergraduate research a few years back; from what I remember, a lot of the identities we wanted to hold formally had additional weird factors on them. Berezinian computations and the resulting identities looking nice depended heavily on wether the exponent in $(-1)^n$ was even or odd in many of the resulting sums over supermatrix indices. | |
| Mar 16, 2021 at 21:41 | comment | added | André Henriques | @NeilStrickland: May I ask you to comment on why the formal properties of the Berezinian are mediocre? | |
| Mar 16, 2021 at 21:22 | history | edited | Nalan | CC BY-SA 4.0 |
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| Mar 16, 2021 at 18:57 | history | edited | Nalan | CC BY-SA 4.0 |
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| Mar 16, 2021 at 17:56 | comment | added | Neil Strickland | For the category of $\mathbb{Z}/2$-graded vector spaces, you can consider the Berezinian (en.wikipedia.org/wiki/Berezinian). The Berezinian is rather complicated and its formal properties are mediocre, despite the fact that the category is very simple. This is not promising for further generalisation. | |
| Mar 16, 2021 at 17:02 | review | First posts | |||
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| Mar 16, 2021 at 16:59 | history | asked | Nalan | CC BY-SA 4.0 |