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bio website math.utoronto.ca/jmfisher
location Toronto
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visits member for 3 years, 6 months
seen Nov 30 at 17:20

I'm a PhD candidate interested in mathematics inspired by theoretical physics (QFT, strings, integrable systems, etc.).


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awarded  Autobiographer
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awarded  Enlightened
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awarded  Nice Answer
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awarded  Yearling
Oct
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answered Reference for Gauss-Manin connection
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awarded  Yearling
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awarded  Revival
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awarded  Enlightened
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awarded  Nice Answer
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comment Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels?
If $K$ is nice, then for each fixed $z$, $K(z,x)$ is a well-behaved function of $x$, hence by the 3rd last equation can express $K(z,x)$ as the sum of $n$ of the integral of $K(z,y) e_n(y)$ with respect to $y$. Now plug in $z = x$ to obtain the result.
Aug
23
revised Help solving differential equation
added 222 characters in body
Aug
23
answered Help solving differential equation
Aug
23
revised Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels?
added 356 characters in body
Aug
23
comment Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels?
Yes, I am well aware and absolutely agree. I answered as above, assuming $K(x,y)$ to be as nice as necessary, since it didn't seem that Mike was asking for the weakest possible hypotheses under which this holds, but rather whether it can be made rigorous at all.
Aug
22
answered Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels?
Aug
21
answered relation between integral and summation
Aug
21
answered Method of characteristics with constant PDE
Aug
13
answered Definition of the rank of the isometry group of a manifold.
Aug
10
comment Are there n-th roots of differential operators?
This technique depends sensitively on both the given differential operator, as well as the space of functions on which it acts. In the case of the Dirac operator, it is not the Laplacian on functions that admits an algebraic square root, but rather the Laplacian on spinors. This also works for vector-valued functions, since we can decompose vectors into products of spinors. These kinds of $n$th roots are completely different than what I described in my answer, but both can be useful depending on the situation.
Aug
10
comment Are there n-th roots of differential operators?
Yes, if you write $\sqrt[1/2]{D_y} \sqrt[1/2]{D_y} g(y)$ as a multiple integral, there will be a term like $\int e^{2\pi i y(k-k')} dy$ which is $\delta(k-k')$ and everything takes care of itself. This is what makes the pseudodifferential calculus work.