Timeline for Proving the basic identity which implies the Chern-Weil theorem
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Jun 25, 2010 at 14:47 | comment | added | Anirbit | Thanks for the details. I have access to the book by Morita. Will look that up. | |
| Jun 25, 2010 at 14:27 | vote | accept | Anirbit | ||
| Jun 25, 2010 at 11:35 | comment | added | Willie Wong | Note that a question which you failed to ask is: even if the identity were tensored against something in $\Lambda^2(T^*M)$, you will still need to define a product between two elements therein for the determinant to make sense. I think you need to go a review some basics: (i) linear algebra over commutative rings [any decent algebra book should do] and (ii) the algebra of differential forms [Morita like I mentioned above, or Madsen & Tornehave's From Calculus to Cohomology also has a description]. | |
| Jun 25, 2010 at 11:29 | comment | added | Willie Wong | Now that addition is defined, we want to also define multiplication. Like I said above, the wedge product extends to a commutative product on even-degree forms. So $\Lambda^*$ forms an algebra, and we can consider matrices with coefficients in it. So in short: you cannot add apples to oranges in either the space of apples or the space of oranges. But if you take the direct sum space, then $\alpha$ apples = $(\alpha,0)$ and $\omega$ oranges = $(0,\omega)$ and you can add them to get $(\alpha,\omega)$. | |
| Jun 25, 2010 at 11:24 | comment | added | Willie Wong | It is rather hard to clarify more. The idea is that the determinant can be defined for a matrix whose coefficients is in some algebraic structure where addition and multiplication is defined. Observe that $\Lambda^{2k}(T^*M)$ is each a vector space. And there are finitely many of them (if $2k > dim(M)$, the space is trivial). So we can form the direct sum $\Lambda^* = \oplus_{k} \Lambda^{2k}(T^*M)$ to obtain a vector space. Each element of $\Lambda^*$ can be decomposed into a sum of a bunch of 2k forms, and when you add two of them you add the ones of the corresponding degrees. | |
| Jun 25, 2010 at 5:54 | comment | added | Anirbit | Thanks for your explanation. But I would have expected the quantity tensored with I as the identity on $\Lambda ^2(T*M)$ instead of a constant function in $\Lambda ^0(T*M)$ as you propose. Can you kindly clarify? (The editor messed up my last comment and neither can I delete it!) | |
| Jun 25, 2010 at 5:52 | comment | added | Anirbit | Thanks for your explanation. But I would have expected the quantity tensored with $I$ as the identity on $\lambda ^2 (T^*M)$ instead of a constant function in $\lambda ^ 0 (T*M)$ as you propose. Can you kindly clarify? | |
| Jun 23, 2010 at 11:19 | comment | added | Willie Wong | ...or rather, the identity transform on the fibre $V$ of $E$ over $M$. | |
| Jun 23, 2010 at 11:17 | comment | added | Willie Wong | Don't treat $H^*$ as a collection of vector-spaces. Instead look at it as a commutative algebra. Let $\mu$ be a 2k-form, and $\nu$ a 2l-form, we can define a product $\mu\nu = \mu\wedge\nu$ which is a 2(k+l)-form. The product commutes for obvious reasons. So the determinant acts on linear transformations of $E$ with coefficients in the commutative algebra of even-degree differential forms. So you should take the identity to be $1\otimes I$, where $1$ is the constant function in $\Lambda^0(T^*M)$ (0-forms, aka functions), and $I$ the identity transform on $E$. | |
| Jun 23, 2010 at 7:00 | comment | added | Anirbit | It is not clear how you can add elements of two different homology classes. They are two different vector spaces! Similar is the confusion with the "+" in the definition of the Chern Form. May be I am missing something obvious. Kindly explain. | |
| Jun 22, 2010 at 22:30 | history | answered | Willie Wong | CC BY-SA 2.5 |