Reputation
20,724
Next tag badge:
96/100 score
20/20 answers
Badges
2 35 78
Newest
 Nice Answer
Impact
~358k people reached

2d
comment Natural isomorphisms: what is the status now of “the Eilenberg/Mac Lane Thesis”?
@Freeze_S, yes, that was my essay about tensor products... True, the story does not take place literally in a category of Hilbert spaces, but in some "comma category" (or whatever people call them these days) of little diagrams from the original category, plus the "new" element of "bilinear maps", and so on. Some people would formalize it more than I did, yes, and some people would even say it's misleading to say things the way I did, but I think that's more about "correctness of formalism" than about underlying reality.
Aug
26
comment Notation for compactly supported functions
Strongly seconding @KeenanKidwell's comment, and your surmise that the subscript $0$ is for functions vanishing at infinity...
Aug
23
comment Does anyone have a proof that the intersection and union of two compact sets is compact.
Conceivably novices to point-set topology could miss @DanielFischer's point: without Hausdorff-ness, compact sets need not be closed! Such stuff! But, yes, with Hausdorff-ness this (possibly counter-intuitive, but relevant to algebraic geometry, at least) issue is skirted.
Aug
23
comment Standard notation for indices in group theory?
@Bernard, I'd hope that "type checking" (i.e., context) in the formal-language sense would easily allow one to distinguish where it was a field extension degree or a group index. And, for my eyes, the plain vertical bars cause far more trouble with my astigmatism than the brackets. Parentheses are not so great, either. :)
Aug
23
comment Standard notation for indices in group theory?
The second one is most standard "in my world". The first one would be intelligible, but cause me to wonder. The third I have never seen used for this purpose.
Aug
20
comment Orbits of the $\text{SL}(n,\mathcal{O}_K)$-action on $\mathbb{P}^{n-1}(K)$ for a number field $K$.
(No, your question is fine for this forum...) The action of $SL_{n+1}(k)$ on $\mathbb P k^n$ is by generalized linear fractional transformations. Given a point in homogeneous coords, first get rid of the denominators to put it in $\mathfrak o^{n+1}$, then use the action of $SL_{n+1}(\mathfrak o)$.
Aug
20
comment Orbits of the $\text{SL}(n,\mathcal{O}_K)$-action on $\mathbb{P}^{n-1}(K)$ for a number field $K$.
@Shadock, :) ...
Aug
19
comment Orbits of the $\text{SL}(n,\mathcal{O}_K)$-action on $\mathbb{P}^{n-1}(K)$ for a number field $K$.
@Shadock "Neither jot nor tittle, dude!" :) Yeah, it's a wee-bit technical. :) But, luckily, there are some techno-tough-people who can protect the civilians from this sort of intellectuo-violence, so, ... well, just salute! :)
Aug
19
answered Orbits of the $\text{SL}(n,\mathcal{O}_K)$-action on $\mathbb{P}^{n-1}(K)$ for a number field $K$.
Aug
19
comment Mathematical formalism for the “dot product” of three vectors
Even without knowing your actual context, probably you want $x\cdot (y \times z)$, that is the dot product of $x$ with the "cross product" of $y$ and $z$. This is a reasonable and useful thing. Also, as it stands, there's sort of a "type error" insofar as I don't know what that alleged ternary operation could mean, since the vectors/matrices are not the right size to be multiplied literally as the notation suggests.
Aug
19
comment Why is the complex number $z=a+bi$ equivalent to the matrix form $\left(\begin{smallmatrix}a &-b\\b&a\end{smallmatrix}\right)$
@RudytheReindeer, you are right that the $\pm b$ can be interchanged without harm, which amounts to switching $\pm i$. My "guessed wrong" was only that I was aiming to "hit" one of the two choices, but got the other one by my choices of convention. Doesn't really matter.
Aug
19
awarded  Nice Answer
Aug
17
comment Does the sum of reciprocals of primes congruent to $1 \mod{4}$ diverge?
@RaceBannon, what does "dare" mean in this context? What sort of game is it? Anyway, this is standard.
Aug
15
answered Weird conformal map problem
Aug
13
reviewed Approve Continuity definition by image rather than pre-image
Aug
12
comment Meaning of $[A,B]$ when $A$ and $B$ are self-adjoint
Aha. I have great respect for the outcomes of Dirac's thinking, but I am not well-acquainted with the literal thing, for various reasons. But/and, knowing what I know of him, he'd have wanted an operation of self-adjoint operators to "return" a self-adjoint operator. The $[P,Q]=1$ issue... Nevertheless, unless I am even more out-of-touch with contemporary physics than I thought I was, this would not be contemporary use, either in mathematics or physics... But I don't claim omniscience... Comment? If you add particulars, perhaps I could add something useful...
Aug
12
revised Finding the Fourier series of a function
edited title
Aug
12
comment Meaning of $[A,B]$ when $A$ and $B$ are self-adjoint
... Oh, and, from your comment, it's not that the vector space of Hermitian matrices is a Lie algebra with that inserted "$i$", although one can insist so, but that skew-Hermitian matrices are, ... with $[A,B]=AB-BA$, as always, ... for the unitary group. I really do think insertion of an $i$ factor is asking for trouble...
Aug
12
comment Meaning of $[A,B]$ when $A$ and $B$ are self-adjoint
More pointedly: I have never in my life seen $[A,B]=i(AB-BA)$, so I'd never suspect the bracket might refer to that. As far as I know, the issue you describe does not actually exist (in any of the several contexts I've seen). It occurs to me that someone might insist that all their operators be self-adjoint, and $AB-BA$ is skew-adjoint for self-adjoint for $A,B$ self-adjoint... and the insertion of the factor of $i$ sorta-artificially makes the outcome again hermitian. So, if anything, the question is perhaps posed oppositely to reported uses? Still, I'd never tack on that bogus "$i$".
Aug
12
comment Meaning of $[A,B]$ when $A$ and $B$ are self-adjoint
I think the point is that for a Lie algebra inside $gl(n)$ there would not be an $i$ as a leading coefficient, ... so there'd be no incompatibility at all. Hermitian matrices are inside $gl(n)$, although not a Lie subalgebra, indeed.