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

As you yourself discovered by finding a paper of Audenaert: an upper bound of the form you require is provided by the Powers–Størmer inequality: Theorem (Powers–Størmer, 1970, Lemma 4.1; link) Let $S$ and $T$ be positive Hilbert-Schmidt operators on a Hilbert space. Then $\Vert S - T \Vert_2^2 \leq \Vert S^2-T^2\Vert_1$, where $\...


12

It is probably dangerous to answer without reading Landsman’s whole paper (and the question seems likely to be closed as “opinion-based”), but I’ll record my first reaction as much like lcv’s (a), namely, it sounds a little bit strawman-ish to separate the two and then pit one (FA) against the other (QM). Footnotes (by Born) on pp. 583, 585 of the famous ...


11

Here is another approach, which like yours is not entirely rigorous, but I think gives a bit more insight into the situation. If you are willing to assume that $f$ is not just smooth but also analytic, then by writing the $f(x+1)$ term as a Taylor expansion you can write the equation in the form $$ (e^D+I) f = g $$ where $D$ is the differentiation operator. ...


11

Pisier https://arxiv.org/abs/1908.02705 very recently constructed a non-nuclear $C^\ast$-algebra $A$ with the weak expectation property (WEP) and the local lifting property (LLP). By a celebrated result of Kirchberg (see Corollary 13.2.5 in Brown and Ozawa's book) it follows that if $B$ has WEP and $C$ has LLP, then $B\otimes_{\max{}} C = B\otimes_{\min{}} C$...


10

About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive. In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above ...


9

No, this already fails in the abelian case. Take $M = L^\infty[0,1]$ with the trace coming from integration against Lebesgue measure. Let $\pi$ take $f(x)$ to $f(x^2)$, i.e., composition with the squaring map. Then $f(x) = x^{-3/4}$ belongs to $L^2[0,1]$ but $\pi(f)(x) = x^{-3/2}$ does not.


9

Assuming that your $\otimes$ denotes min-tensor product of ${\rm C}^\ast$-algebras, then the answer to the question in the body of your post -- which is NOT the same as the question in the title of your post, at present! -- is "Yes, but this says nothing special about $K$." In more detail: min-tensoring an inclusion of ${\rm C}^\ast$-algebras with any fixed ...


9

The solution $Z(t)$ of your differential equation with $Z(0) = Z_0$ satisfies $$ Z(t) (e^t + (1-e^t) Z_0) = Z_0 $$ In order for this to be periodic with period $p$, you'd need $(1-e^p) Z_0 (1-Z_0) = 0 $. $1-e^p = 0$ (for real $p$) only if $p=0$, while if $Z_0 (1-Z_0) = 0$ we have a fixed point.


9

It seems to me that the fissures between (sub-)disciplines are somewhat more complex than the simple functional analysis vs. quantum theory dichotomy that Landsman emphasizes. The study of foundational questions in physics is pursued by a comparatively small group of researchers. After all, physics has to deal with observations of the real world, and the ...


9

The answer is no, in general. Here is a counterexample: Let $A$ be the algebra of bounded linear operators on $\ell^2(\mathbb{N})$, and let $a \in A$ be the left shift on $\ell^2(\mathbb{N})$. Then the spectrum of $a$ is the closed unit disk $\overline{D}$, and the point spectrum of the operator $a$ is the open unit disk $D$. Now we note that the notion "...


9

Yes to both.$\newcommand{\Cst}{{\rm C}^*}$ The standard example for the first is: take a discrete group $G$ and let $A$ be its full $\Cst$-algebra, $B$ its reduced $\Cst$-algebra. There is a canonical homomorphism $q:A\to B$ which is injective when restricted to $\ell^1(G)$; but $q$ is injective if and only if $G$ is amenable. So any non-amenable discrete ...


8

By "trace" I assume you mean tracial state, and in that case the answer is "not necessarily". A counter example is produced below. The goal is this: we construct an unital RFD $C^\ast$-algebra $B$ and a $\ast$-epimorphism $\pi \colon B \to \mathcal O_2$ (Cuntz algebra). Then $A := \ker \pi$ does the trick. In fact, it is RFD since this property is preserved ...


8

The answer is yes. An easy way to see this is to first check that the inclusion of $M_n(A)$ into $M_n(B)$ is isometric. Then taking the union over $n$ yields the result (whether you complete or not, since it's isometric on the uncompleted union).


8

For $\xi,\eta\in H$ let $\theta_{\xi,\eta}$ be the rank-one operator $\theta_{\xi,\eta}(\gamma) = (\gamma|\eta) \xi$ for $\gamma\in H$. Let $(e_i)$ be an orthonormal sequence in $H$, set $S_i = \theta_{e_1, e_i} \in B(H)$ and let $R_j$ be the projection onto the span of $\{e_1,e_2,\cdots,e_j\}$. Then $$ \lim_i \lim_j (S_iR_j(e_i)|e_1) = \lim_i (S_i(e_i)|...


8

There are even commutative counterexamples. Let $A = C[0,2]$ and let $A_0$ be the $*$-subalgebra of all polynomials in $x$. Then let $p: C[0,2] \to C[0,1]$ be the restriction map. (My first example took $A = C[0,3]$ and let $A_0$ be the set of all polynomials in $x$ with rational coefficients and $p: A \mapsto \mathbb{C}$ the point evaluation at $x = e$. ...


7

See JMG Fell, The Dual Spaces of C$^*$-Algebras, Trans. Amer. Math. Soc. 94 (1960), 365-403, Corollary 1 on p. 388: Let $A$ be a C$^*$-algebra with dual space $\hat{A}$. Let $T^i$ be a net of elements of $\hat{A}$, all of dimension equal to or less than the integer $n$; and let $S^1, \ldots S^r$ be distinct elements of $\hat{A}$ such that $T^i\to_i S^k$ for ...


7

Let $C$ denote the complexes, and embed $C \times C$ in $M_2 C$ ($2 \times 2$ matrices) as diagonal matrices. Then $M_2 C$ has unique trace, but $C \times C$ has two extremal ones.


7

It is not Type I in general. Probably it is not Type I whenever $G$ is infinite. Here is an argument when $G=\mathbb{Z}.$ Consider the projections in $\ell^\infty(\mathbb{Z})$ defined by characteristic functions for the following sets $\{ 2^n\mathbb{Z}+k: n\geq 1,0\leq k <2^n \}.$ Let $A$ be the C*-algebra generated by these projections. Then $A=C(X)$...


7

I know my answer is coming a bit late, but the answer to your question is: yes. If $A$ is a $C^\ast$-algebra, and there exists a $C^\ast$-algebra $B$ such that $A\otimes_\alpha B$ is strongly Morita equivalent to $\mathbb C$ for some $C^\ast$-tensor product $\otimes_\alpha$, then $A\cong \mathcal K(H)$ is the compact operators on some Hilbert space $H$, and ...


6

Assuming complex scalars, no, any inner product which satisfies (2) for all $v \in \mathcal{V}$ and $w \in \mathcal{W}$ has the given form. To see this, let $[\cdot,\cdot]$ be any inner product which satisfies (2). So right away we know that $[v\otimes w,v\otimes w] = \langle v\otimes w, v\otimes w\rangle$ for all $v$ and $w$. Next, for $v,v' \in \mathcal{V}$...


6

There are a number of ways to do this calculation, but at risk of shamelessly plugging my own work there is a nice way to see it using a Mayer-Vietoris principle. Decompose $\mathbb{R}$ as the union of the rays $\mathbb{R}^+ = [0, \infty)$ and $\mathbb{R}^- = (-\infty, 0]$, intersecting at a point. This decomposition admits Mayer-Vietoris sequences in both ...


6

Since questions recirculate to the front page forever if left unanswered, I will amalgamate the comments into an answer, which I've made community wiki. Firstly, by the characterization of the Cantor space as the only metrizable Stone space with no isolated points (up to homeomorphism), $\newcommand{\Z}{\mathbb{Z}}\Z_p$ is homeomorphic to the Cantor space. ...


6

I think this is just a definition chase, and use of the standard representation of a trace-class operator. Starting with the later, following Chapter II in the first volume of Takesaki, for example, we have that any trace-class operator $a$ can be written as $$ a(\xi) = \sum_n a_n (\xi|\xi_n) \eta_n \qquad (\xi\in H), $$ where $(\xi_n),(\eta_n)$ are ...


6

I'm not sure what you mean by "quantum integer space", but if you're asking for an analog of the Fourier correspondence between $\mathbb{T}^2$ and $\mathbb{Z}^2$ I think the answer is no. We do have a notion of duality for quantum groups, but there is no reasonable way to put a quantum group structure on the quantum torus.


6

Well, the space of Hilbert-Schmidt operators is a Hilbert space, so you are asking whether weak convergence to zero implies norm convergence to zero in a Hilbert space. The answer is no. For instance, let $A_n$ be the rank 1 projection onto $e_n$. This converges weakly but not in norm to zero.


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