38

I will have another go at arguing that dagger-categories are not evil.
Let’s look at a simpler case first. Consider the property “$1 \in X$” on sets. As a property of abstract sets, this is evil: it’s not invariant under isomorphism, e.g. any iso $\{1,2\} \cong \{2,3\}$.
But it is manifestly non-evil as a property of, say, “sets equipped with an ...

answered Oct 5 '15 at 22:33

Peter LeFanu Lumsdaine

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John Pardon's invariant version of $\dagger$-category is very good, but unfortunately it does not capture the examples. Specifically the category of Hilbert spaces doesn't have this structure.
There is a problem in making the isomorphisms such as $s \circ \dagger_* \cong t$ natural. Such a natural isomorphism gives component morphisms
$$\phi_y: y^\dagger ...

answered Oct 5 '15 at 14:03

Chris Schommer-Pries

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20

The answer is yes:
Let me sketch the proof. So $p:E\to M$ is the fiber bundle with typical fiber $F$ which is compact, connected (and oriented, for simplicity's sake), and you are given a vertical volume form $\mu$; so $\mu_x$ is a volume form on each fiber $E_x$ which depends smoothly on $x\in M$. First I choose another vertical volume form $\nu$ such ...

20

In my opinion, dagger categories are not evil, but they are not categories with extra structure (in the ordinary sense) either. (I think this is what Qiaochu Yuan meant in his comment.) More specifically, I wouldn't think of $\dagger$ as a functor. Let me give a comparison.
In some parallel universe, people started to study "ategories", before they studied ...

16

One potential intuition for the trace norm is as a way of turning the rank of a matrix (which is very discontinuous) into a norm (which is continuous). Specifically, the trace norm is the unique norm with the property that $\|P\|_{\mathrm{tr}} = \mathrm{rank}(P)$ for every orthogonal projection $P \in M_n(\mathbb{C})$.
Closely-related to this is the ...

answered Aug 6 '17 at 0:49

Nathaniel Johnston

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From four subspaces in general position one can generate an infinite number of other subspaces by closing up under joins and meets. This is true even for subspaces of $\mathbb{R}^3$ (any field of characteristic zero in place of $\mathbb{R}$ would do). This is easy to see in the corresponding projective plane picture: take $a = (0: 0 : 1)$, $b = (1: 0: 1)$, $...

answered Aug 8 '15 at 3:25

15

As I Said in a comment above, Peter Selinger posted a few years ago on the category theory list a definition of a "non-evil structure" and showed that there is no reasonable way to make the notion of dagger categories into such a notion. His post can be found here.
The core of the argument, as I understand it, is that if you have an equivalence of category $...

14

By integration by parts,
$$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \int x^2 f'' f.$$
$$-\int 2xf'f = \int f^2=\|f\|^2.$$
By the Cauchy-Schwartz inequality
$$\bigg|\int x^2 f'' f\bigg|\le \|x^2f\|\|f''\|.$$
The conclusion is proved.

fa.functional-analysis real-analysis ap.analysis-of-pdes ca.classical-analysis-and-odes hilbert-spaces

13

To the contrary, my feeling is that nonseparable Hilbert spaces are in some sense artifacts and can almost always be avoided. And more generally, to your comment about nonseparable Banach spaces being "nothing special", most of the nonseparable Banach spaces one meets in practice are duals of separable Banach spaces, and therefore are weak* separable. ...

13

Another answer is that $M_n$, the space of $n\times n$ complex matrices, carries an operator norm where the norm of a matrix is its norm as a linear operator from $\mathbb{C}^n$ to itself (giving $\mathbb{C}^n$ euclidean norm). For some of us, this is the most natural and useful norm on $M_n$.
With operator norm, $M_n$ is a finite dimensional Banach space, ...

11

There are many examples constructed with weighted shifts. The following is a Hilbert space example.
Let us consider the Hilbert space $L^2\big((0,1),\mu\big)$, where $\mu$ denotes the measure defined by
$$\mu(A):=2\lambda(A\cap(0,\tfrac12))+\lambda(A\cap(\tfrac12,1)).$$
for all Lebesgue measurable sets A. Here $\lambda$
is the Lebesgue-measure. Furthermore, ...

11

"The" completion is not always a space of functions, for $N=1$ or $2$ for example it is a quotient $D^{-2}L^2/P_1$ (equivalence classes of functions $u\in H^2_{loc}$ with $\partial_i \partial_j u\in L^2$ with $u\sim v$ iff $u(x)-v(x)=a\cdot x+b$.
For $N=3$ and $4$ it is $D^{-1}\dot{H}^1/P_0$ where the "homogeneous Sobolev space" $\dot{H}^1$ is "the" ...

11

I don't see which kind of condition you are looking for, as there are a lot of pairs $T,S$ such that $TS=\lambda ST$ and $\lambda\ne1$, even in finite dimension. Such pairs are said to $\lambda$-commute.
An interesting case happens when $\lambda$ is root of unity, say of order $r$. Then (Potter's Theorem) $(T+S)^r=T^r+S^r$.
Still when $\lambda^r=1$, here ...

answered Feb 26 '18 at 13:06

Denis Serre

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11

Every separable Banach space is a quotient of $\ell_1$, so in particular every subspace of $\ell_1$ is a quotient of $\ell_1$.

11

By a cutoff function argument, it suffices to assume $f$ is compactly supported, so we can integrate by parts without picking up boundary terms.
Thus
$$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \int x^2 f'' f$$
Hence using Cauchy-Schwarz,
$$\|xf'\|_2^2 \le \int |2xf f'| + \int |x^2 f f''| \le 2\|xf\|_2 \|f'\|_2 + \|x^2f\|_2 \|f''\|_2.$$
The second term ...

fa.functional-analysis real-analysis ap.analysis-of-pdes ca.classical-analysis-and-odes hilbert-spaces

10

This is not an answer but rather a comment to Peter Michor's answer. Anyway, I post it as an answer to get more flexibility in text formulation and to get more visibility. Namely, I think there is a crucial error which completely breaks down the argument so that generally it is not possible to perform the construction (2• of OP) of associating a ...

10

Yes, there are such tilings, but I don't know of any "nice" ones. The example below makes use of the axiom of choice, so you can guess how horrible it is. :)
Take any dense countable subgroup $J$ of $H$ - say, the subgroup of vectors with rational coordinates, such that finitely many of those are nonzero. Now for every coset $x + J$ there is a ...

10

In the definition, "diagonal" does not make much sense, you probably mean "self-adjoint", although this can be relaxed to "m-sectorial" (powers of $A$ still make sense then). Also, it is irrelevant for the definition whether or not $A$ has some essential spectrum.
Anyway, the main reason why these interpolation spaces are useful for the analysis of SPDEs (...

fa.functional-analysis ap.analysis-of-pdes operator-theory hilbert-spaces stochastic-differential-equations

10

General fact: if we remove a countable collection $K_1,K_2,\ldots$ of compact sets from an infinite-dimensional Banach space $X$, the remaining set $V$ is locally and globally path-connected (actually any open ball is path-connected).
Proof. If $K$ is a compact subset of $X$, and $x\in X$ is a point, then the set union $\cup_{z\in K} xz$ of segments $xz$ is ...

answered Feb 26 '15 at 22:36

Fedor Petrov

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10

OK, let me try too. It is going to be a somewhat long story. WLOG, $\|T\|\le 1$.
Step 1: It is enough to show that for every finite-dimensional subspace $E$ and every $\delta>0$, there exists a unit vector $v\in E^\perp$ and $z\in\mathbb C$ such that $\|Tv-zv\|\le\delta$.
Proof: Suppose that the claim holds. Choose a square summable sequence $\delta_j&...

10

We can tell $\mathbb{R}$ and $\mathbb{R}^2$ apart by topology, and we can tell $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^2)$ apart by quantum topology. In the sense that a choice of C*-algebra contained in $B(H)$ amounts to a choice of quantum topology on $H$.
In your case that C*-algebra would be either $C_0(\mathbb{R})$ or $C_0(\mathbb{R}^2)$, realized as ...

9

Let $P=\left[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right]$ and $Q(\phi)=\left[\begin{matrix} \cos^2(\phi) & \cos(\phi)\sin(\phi) \\ \cos(\phi)\sin(\phi) & \sin^2(\phi) \end{matrix}\right]$. Then $P$ and $Q$ are orthogonal projections and if $\phi\neq 0$, then the only operator $T$ which satisfies $0 \leq T \leq P$ and $0 \leq T \leq Q(\...

9

Such a net exists for any $\Phi$ (in fact, there is a canonical such net). First, note that if $F\subseteq V$ is a finite-dimensional subspace, then there is a unique $\phi_F\in F$ such that $\Phi(\psi)=(\phi_F,\psi)$ for all $\psi\in F$. The collection of such $F$ form a directed set under inclusion, and the net $(\phi_F)$ will have the desired property.

9

It's not true.
Edit: thinking twice there's a simpler solution.
On $L^2([0,1])$, define $f(x)(t)=1+t^{1/2x}$ for $x\in\mathopen]0,1]$ and $f(0)=1$. It's free (because the $t\mapsto t^x$, $x\ge 0$, form a free family). It depends continuously on $x$ (check at $x=0$). For $x\le 1/2$ it contains all (restrictions of) polynomials hence spans a dense subspace. ...

9

As noted in comments, there are no bounded operators whose commutator with their adjoint is a non-zero scalar. The prototype of unbounded operators with the desired property is the 1-dimensional Dirac operator $T=ix\pm i{d\over dx}$ on $L^2(\mathbb R)$. [Correction, thanks to @ChristianRemling for noticing my blunder!) The composition $T\circ T^*$ (not ...

8

This is stated and proved as Proposition 2.20 here:
https://isem-mathematik.uibk.ac.at/isemwiki/images/9/94/ISEM15_Lecture2.pdf

8

For $p\neq 2$ there are plenty of extreme points in $K(\ell_p)$.
J. Hennefeld, Compact extremal operators, Il. J. Math. 21 (1977) 61-65.

8

The surprising fact is that the GNSStinespringKasparov theory is in fact completely algebraic, at least to a very large extend: the following results have been obtained by a PhD of mine but are, unfortunately, not published (yet?).
So the setting is a *-algebra over a ring $C$ which is of the form $C = R(i)$ with $i^2 = -1$ and $R$ being an ordered ring. ...

8

As requested, I've moved my comments to an answer.
The question is equivalent to the following:
Suppose that a doubly-infinite matrix $A=(A_{ij})_{i,j\in {\bf Z}}$ represents a bounded hermitian operator on $\ell^2$. Does the matrix $(|A_{ij}|)$ also represent a bounded operator on $\ell^2$ ?
(The original question just considers matrices indexed by ${\...

8

Let $X$ be a Banach space such that $X^*$ is separable. I claim that the space of compact operators $\mathcal K(X)$ on $X$ is separable.
It is enough to show that $nB_{\mathcal K(X)}$, the space of compact operators of norm at most $n$, is separable as then $\mathcal K(X) = \bigcup_{n=1}^\infty nB_{\mathcal K(X)}$ is separable too.
By Shauder's theorem, ...

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