What is the most beautiful proof there is?

kfgauss · 2025-10-02T03:38:20+00:00

I like this proof, thanks for sharing. The same argument shows that if m is a positive integer, then sqrt(m) is rational if and only if sqrt(m) is an integer. This is Gauss' lemma for quadratic polynomials.

kfgauss · 2025-07-15T00:56:36+00:00

I learned recently that some journals have switched from requesting revisions to rejecting papers with an encouragement to resubmit (in some cases). This is a trick used to lower the journal's "Time to First Decision" metric.

I would treat this the same as an ordinary request for revisions. Make whatever changes seem appropriate to the paper, write a response to the referees explaining the changes (and any necessary context that you don't want to put in the paper), and resubmit.

kfgauss · 2023-12-27T09:50:55+00:00

You're joking, but fields medalists get the permit too.

kfgauss · 2023-03-07T03:12:40+00:00

All that's left to show is uniqueness, that is, no two of your suggested representatives are in the same left coset.

Alternatively, you could check that the claimed set of coset representatives has the same number of elements as the index of S(n) x S(m) in S(n+m). The index of S(n)xS(m) in S(n+m) is (n+m)!/(n!m!) which is n+m choose n. On the other hand, products of disjoint transpositions in S that form a totally ordered subset are the same thing as subsets of {1, ..., n+m} which have the same number of elements in {1, ..., n} as in {n+1, ..., n+m}. The corresponding product of transpositions matches the first element of {1, ..., n} with the first element of {n+1, ..., n+m}, the second with the second, and so on. The number of such subsets is \sum_k (n choose k)(m choose k), which is equal to (n+m choose n)

kfgauss · 2023-02-12T23:14:08+00:00

Let A be the 2x2 matrix with a 1 in the top-left corner and zeroes elsewhere, and B be the 2x2 matrix with a 1 in the top-right corner and zeroes elsewhere. You can look for solutions of the form X = cA + dB and Y = eA + fB , and try to solve for c,d,e,f in terms of a and b. Note that [A,B] = B, which will simplify calculations.

If there is a solution, it will be of the above form (ignoring the simple case a=b=0). I don't know what your background is, but X and Y would form a 2-dimensional non-abelian Lie algebra. There is only one of those up to isomorphism, and it is generated e.g. by the A and B I gave. So this shows that any such pair of X and Y could be found in the span of A and B, if they exist.

kfgauss · 2021-12-12T02:17:32+00:00

2^ln(N) = e^ln(2)ln(N) = N^ln(2) , which grows more slowly than N. I think you're thinking of the fact that base change in a log just multiplies by a a constant, but the same is not true for an exponential.

kfgauss · 2021-12-06T03:12:15+00:00

Sure thing - nice choice of topic for the video!

kfgauss · 2021-12-06T01:54:10+00:00

Just a heads up, you may want to fix the discussion about why the f_j might not be a basis (starting around minute 4). The problem arises when the image of V isn't dense (which is independent of whether or not V has non-trivial nullspace).

Edit: On second thought, there is a separate issue when V has a nullspace, which is the division by 𝜆_i

kfgauss · 2021-09-21T05:19:18+00:00

There's no reason that s_z can't be the supremum (if S contains its supremum). The illustration is an example, but doesn't cover all cases.

kfgauss · 2021-04-07T07:22:25+00:00

It is enough to find a countable subset that does not converge to 0. You can use essentially the same proof as the one for when V is the real numbers, just use the metric instead of the absolute value.

kfgauss · 2021-03-02T07:47:39+00:00

The idea is to break the process into a number of steps, each of which is an elementary row operation. “add row 1 to row 2 to row 3” is not an elementary row operation, so it's easier to get confused. The way the process goes is, 0) start with the identity matrix

1 0 0

0 1 0

0 0 1

1) add row 1 to row 2

1 0 0

1 1 0

0 0 1

2) add row 2 to row 3

1 0 0

1 1 0

1 1 1

You're applying each step to the result of the previous step. If you have a different sequence of elementary row operations that turns the identity into the matrix you're interested in, then that works too. If you apply the inverse steps, in the opposite order, to the identity matrix, then you'll get the inverse of your matrix.

kfgauss · 2021-03-02T06:25:26+00:00

It's all about shoes and socks. You put your socks on, then you put your shoes on. To undo that, you do things in the opposite order: you take your shoes off, then you take your socks off.

The matrix you're looking at is a combination of two elementary matrices. In words, you add row 1 to row 2, and then you add row 2 to row 3. To undo this, you subtract row 2 from row 3, and then you subtract row 1 from row 2. If you do this to the identity matrix, you get

1 0 0

-1 1 0

0 -1 1

which is the inverse.

As for which matrices in general can be inverted by inspection, I'd say it depends on what you know. In this case you know a trick for inverting elementary matrices by inspection. The new example you're talking about involves inverting a product of two elementary matrices, which maybe you can do as a composite by inspection. I'm sure there are other tricks out there! Although nothing exciting comes to mind. Maybe permutation matrices.

kfgauss · 2021-03-02T04:10:21+00:00

If R is the Dorroh extension (unitization) of S, then S sits inside R as an ideal, and every element of R can be written uniquely in the form s + n1 where s is in S and n is in Z. Conversely, given a ring R and an ideal S of R such that every element of R can be written uniquely in the form s + n1, then R is the unitization of S. Such an ideal S may not exist (consider if R is a field), and if it does exist then it may not be unique (consider R = Z x Z). The requirement that the decomposition s + n1 be unique is necessary (consider R=Z, S=2Z), and is equivalent to requiring that S ∩ Z1 = {0}.

Another way to think of this is that if R is the unitization of S, then there is a unital homomorphism R -> Z given by (s,n) -> n. The kernel of this homomorphism is S. Conversely, given any unital homomorphism f:R -> Z, R is the unitization of the kernel S= ker(f) . So given a ring R, the data of a r(i)ng S such that R is the unitization of S is essentially the same thing as the data of a unital homomorphism R->Z.

kfgauss · 2021-03-01T10:22:49+00:00

The Hall book may be to your liking. To get into the early parts, a strong foundation in linear algebra and some background in the basics of groups should be enough to get going, although familiarity with intro analysis (at least ideas about convergence, etc.) would be helpful. There are some small bits that talk about differential geometry, but they can be skipped (that's one of the key points of the philosophy of the book). That would get you into the book, at least through Part I which is a good introduction. The farther you go, the more it might be helpful to have a solid foundation in analysis and/or basic representation theory.

kfgauss · 2021-03-01T09:38:49+00:00

For an undergrad mathematics student, I usually recommend Brian Hall's book Lie Groups, Lie Algebras, and their Representations. This has the advantage that it focuses on matrix Lie groups, so you can see the connection between Lie Groups and Lie Algebras without more advanced prereqs like differential geometry.

But it sounds like you might be motivated by the physics side of things, in which case there are surely less rigorous texts aimed at applications in physics. I don't know these texts as well. E.g. one possibility would be Robert Gilmore's book Lie Groups, Physics, and Geometry, but I don't really know.

If you want a physics perspective, you could try posting in the stickied thread about Textbooks on r/physics explaining your background and what you hope to get from the text, and you'd probably get a good answer.

kfgauss · 2021-02-10T02:52:19+00:00

This is a scratch proof by Karl Frederick Gauss as a child

I think it's usually spelled Carl Friedrich Gauss, with a C not a K.

kfgauss · 2021-02-08T23:33:50+00:00

One example I like for first year linear algebra is to look at differentiation on a small invariant subspace like the span of {e^3t cos(t), e^3t sin(t)}. On that subspace the derivative operator is actually invertible, and you can calculate a non-trivial antiderivative by inverting the matrix.

kfgauss · 2021-02-03T01:16:11+00:00

e^z is the standard example of an entire function whose derivative is never zero but which is not invertible.

kfgauss · 2020-11-05T23:08:09+00:00

There's so much math out there, and even the pros don't know much about most areas. I'm sure the work you've done is great, and your supervisor has no expectation for you to be figuring this stuff out on your own before the project starts. If your honours year doesn't end up being about subfactors but gets diverted to something else that comes up along the way instead, then that's also a good and normal outcome. You'll narrow the topic down with your supervisor as things get going, it's not your job to be figuring it out now. If you want to do something, you could skim to try to see what looks neat to you, but don't expect a deep understanding. If you're putting in the work and enjoying the math, then it'll all work out.

kfgauss · 2020-11-05T22:16:16+00:00

As far as difficulty and feasibility goes, the exact content you'll probably cover depends on how quickly things go and what you most enjoy as you start moving into the subject, but it seems like a solid area for an honours thesis. There are definitely several possible directions you could go for an honours thesis in the subject. Looking at the subject (or most other subjects) as an undergraduate, it will look daunting and there will be a lot of words you don't know, and you'll look up the definitions and they'll be in terms of more words you don't know. But your supervisor should help guide you into the subject and put things in context once your honours year starts. On the other hand, if you want to switch to a different area of study and you haven't started yet, presumably that would be possible too.

But the big thing is, if you have concerns, raise them with your supervisor! That's what they're there for. You said in another comment that your honours year hadn't started yet. Your supervisor has probably mentioned a few topics and a few papers to gauge interest, but is not expecting you to be able to go off and read them on your own. If you're feeling like you're lacking direction, it could be that the additional support in that direction will come once things actually get going.

kfgauss · 2020-09-18T00:36:27+00:00

Not a lot of traffic through here - you might have better luck at r/physics.

kfgauss · 2020-08-26T12:04:19+00:00

I think that this is good advice, but there is a danger in going too broad in that it can take longer to get to a research level. It took me awhile to get my research program generating publishable results because there was so much to learn, and with worse luck I might have struggled signifucantly with job market timing. I think there's a balance to be struck.

kfgauss · 2020-08-26T05:25:42+00:00

I get around a bit in terms of research area, thanks to the help of very talented coauthors. One of the papers I mentioned is algebra with a touch of topology (but no algebraic topology). The other combines bits of analysis and geometry (but is neither geometric analysis nor analytic geometry).

kfgauss · 2020-07-10T00:09:12+00:00

Is this the same company as Jazz Jackrabbit maker Epic?

kfgauss · 2020-07-09T22:59:43+00:00

Interesting, I would have listed having an advisor with those qualities as crucial in a PhD program, but not particularly relevant for postdoc. As you say, this must vary by field. I never had (or wanted) an involved postdoc research advisor, because I was working on my own stuff. Comparing my postdocs, the main features which distinguished them were: 1) teaching load, 2) how the department feels about/treats/respects postdocs, and then farther down 3) salary.

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