Timeline for The trace of a wedge product of matrices
Current License: CC BY-SA 3.0
12 events
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| Mar 16, 2020 at 7:24 | comment | added | Guest123412341234 | It seems like the implied norm is defined so that $|W^+|\omega_g=-\frac12\operatorname{tr}(W^+\wedge W^+)$. I don't see how this is necessarily well defined. ie Why is $\left<\operatorname{tr}(W^+\wedge W^+),\omega_g\right>\leq 0$? | |
| Feb 21, 2018 at 20:19 | comment | added | Robert Bryant | @BrianKlatt: There's no particular advantage. It's just that, if you happen to know an example in which all the calculations have been done and you can evaluate the constant by quoting that example, then it gives a quick way to get the answer. If you don't know the integral, etc., then it's probably better to just calculate algebraically at a point and follow the definitions, always being careful that you are using the standard normalizations for norms of tensors, etc. | |
| Feb 21, 2018 at 20:08 | comment | added | Brian Klatt | @RobertBryant One last question: Is there any advantage to computing the constant $c$ by doing the integral over $\mathbb{CP}^2$ rather than just doing a pointwise computation using a curvature tensor whose only nonvanishing component is $W^+$ (or is there something conceptually wrong with doing the latter)? | |
| Feb 21, 2018 at 17:47 | comment | added | Brian Klatt | @RobertBryant Oh, it's just Schur's Lemma then, thanks. | |
| Feb 21, 2018 at 8:17 | comment | added | Robert Bryant | @BrianKlatt: Yes, it's a consequence of the representation theory. No 'cross term' could be invariant under change of frame, so it can't appear. The point is that, if $V$ and $W$ are non-isomorphic irreducible representations of a compact group $G$, then there cannot be a nonzero $G$-invariant bilinear mapping $V\times W\to \mathbb{R}$, because then you'd have an isomorphism $V\to W^*\simeq W$. | |
| Feb 21, 2018 at 4:19 | comment | added | Brian Klatt | Any chance you could say a brief word about why the universal quadratic expression is a priori a sum/difference of squares coming from the irreducible components and doesn't involve "mixed" terms, say, inner products between different irreducible components? I presume it's immanent in the representation theory but I'm a rube on the subject at the moment (and for the foreseeable future). | |
| Mar 8, 2014 at 21:37 | vote | accept | Geom math | ||
| Mar 5, 2014 at 19:45 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added 123 characters in body
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| Mar 5, 2014 at 19:37 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added some information
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| Mar 5, 2014 at 19:05 | comment | added | Robert Bryant | Sure. I know `why' the sign appears and why $B$ and $B^*$ don't. Briefly, the sign in front of $|W_+|^2$ must be negative because $\mathbb{CP}^2$ with its canonical orientation is self dual. And $|B|^2$ can't appear in the formula for equivariance reasons (I'll add a few words of explanation about this in the answer above). | |
| Mar 5, 2014 at 18:41 | comment | added | Geom math | I understand your explanation, I think the 2 factor appears because the basis is just ortogonal (It's necessary divide by $\sqrt{2}$ the terms of the basis to obtain a ortonormal basis),but do you know why the "-" sign appears, in my mind the sign is the opposite. Another question is why the terms $B$ and $B^*$ don't appears in the result? The Besse's explanation is not clear to me, more specifically, why $R\wedge R$ annihilates the terms $B$ and $B^*$? | |
| Mar 5, 2014 at 12:54 | history | answered | Robert Bryant | CC BY-SA 3.0 |