# Tag Info

47

I'm can't claim to have studied the relevant history in a lot of detail, but count me a skeptic of Landsman's claim. Let's take this little paper and the companion that it cites as a test case, which I hope we can all agree is "real physics". The authors are clearly well versed in the calculus of variations and the representation theory of Lie groups. ...

26

This reminds me the following anecdote. K. Friedrichs once met Heisenberg on a conference. He thanked Heisenberg for creation of quantum mechanics which benefited mathematics so much, and added: "But mathematicians gave much in return." Heisenberg: "What?" Friedrichs: "For example, von Neumann explained the difference between symmetric and self-adjoint ...

18

A direct and precise way to arrive at your answer is to appeal to the theory of Fourier transforms of distributions. If I define the Fourier transform as $$F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{ikx}\,dx,$$ then the equation $f(x+1)+f(x)=g(x)$ becomes upon Fourier transformation $$e^{-ik}F(k)+F(k)=G(K)\Rightarrow F(k)=\frac{G(k)}{1+e^{-ik}}.$$...

18

As jjcale mentions in a comment, the index of a Fredholm operator is very important in physics. One way to define the Chern number of a topological insulator is in terms of the index of a Fredholm operator, as explained in [1]. There is also the concept of an index of a pair of projections. This is seen a lot recently in physics papers, for example in [2]. ...

14

Yes: A $C^*$-algebra satisfies the identity $e^{[xy-yx]}=e^xe^ye^{-x}e^{-y}$ iff it is commutative. This follows from two independent facts (I write $[x,y]=xy-yx$) 1) A (real/complex) unital Banach algebra satisfies the identity $e^{[xy-yx]}=e^xe^ye^{-x}e^{-y}$ $\Leftrightarrow$ it satisfies the identity $[x,[x,y]]=0$ $\Leftrightarrow$ it satisfies ...

12


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