Tweet at me your favorite differential operator and why you think it's better than the Hodge Laplacian
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Replying to @duetosymmetry
The Dirac operator, because it detects inequivalent spin structures. Plus, a curly d with a slash through it looks really cool.
1 reply 1 retweet 21 likes -
Replying to @johncarlosbaez @duetosymmetry
Would it not be more accurate to say that the Dirac operator is meaningless (i.e., undefined) until and unless a particular spin structure is specified? (Absolutely agree, of course, about the attractiveness of the notation.)
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Replying to @MathPrinceps @duetosymmetry
If you changed "more accurate" to "true" I would agree whole-heartedly.
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Replying to @johncarlosbaez @duetosymmetry
Then would you please clarify what you meant when you said that the Dirac operator can distinguish between spin structures? Is it merely that the Dirac operator is, in a suitable sense, a one-to-one function of the spin structure?
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Replying to @MathPrinceps @duetosymmetry
No, I meant something weaker and well-known: the spectrum of the Dirac operator depends on the spin structure, so we can *sometimes* use it to distinguish between spin structures: https://arxiv.org/pdf/math/0007131.pdf … Btw, I thought Leo's question was a joke, so my response was a joke.
3 replies 0 retweets 4 likes
Thank you. I am embarrassed to confess that I was unaware of this work, but now that I see it, of course it all seems very natural and unsurprising.
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Replying to @MathPrinceps @duetosymmetry
No need to be embarrassed - there's way too much math for anyone to know all of it!
0 replies 0 retweets 3 likesThanks. Twitter will use this to make your timeline better. UndoUndo
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