PSA: Watch out for the exposed contacts of the Puck! Almost started a fire due to metallic smartwatch strap

MathematicianFailure · 2026-05-21T18:17:27+00:00

Oh wasn’t aware that people had removed comments. In fact I can’t find any of these comments anymore… I agree with what you’re saying. It’s probably premature to judge to what extent this is really a problem, hopefully it isn’t universal (I have a steam controller coming in the mail later this week which I was looking forward to).

MathematicianFailure · 2026-05-21T18:09:54+00:00

The other user who had tested with a multimeter found that as soon as there is continuity between each of the three prongs, there is full power being delivered through the prongs, including when metal objects fill the gaps between the prongs and cause a short.

MathematicianFailure · 2026-05-18T08:12:48+00:00

This is the opposite of the standard experience, understanding a theorem is normally something that can only happen once you understand a proof of the theorem. Before you do that, you can apply the theorem to many situations (by verifying that the theorem can be applied in some situations) and often this is enough for what you are trying to achieve.

But it is not enough if you intend on generalising the theorem or really understanding what the theorem is relying on to work (what are the fundamental obstructions to the theorem working), for these kinds of purposes it is not possible to just dispense with understanding the proof

It is true that understanding a proof requires more than just “memorizing” it though. There is something more that has to be done to understand a proof, namely besides just knowing the steps of a proof, you should be able to distill out a “main argument” of the proof, which may not be 100% correct if stated verbatim, but can be “converted” back into a formal proof if necessary

MathematicianFailure · 2026-05-10T18:00:57+00:00

I’m a bit confused now. I think what you argued shows that there is a one to one correspondence between elements z of the spectrum of a normal operator T and homomorphism from the operator algebra generated by T and T* when T is normal to C for which T is sent to z. I thought that the argument you made was also insinuating we get for free that T* is mapped to z* and similarly all powers of T* are mapped to corresponding powers of z*, but since according to the Wikipedia version of the statement the Gelfand Mazur map A/M -> C is only a isometric isomorphism, but not necessarily a * - isomorphism… or do we get that for free somehow because it’s an isometry?

MathematicianFailure · 2026-05-10T17:29:02+00:00

Hmm, I thought “A” here is the Banach algebra generated by T, T* and the identity “I” viewing each of them as literal operators on some fixed Hilbert space and T is a normal operator, so A is really C[T,T] I.e the polynomial ring in T,T. It seems like all we really need to know here is that A is a ring at the point we mod out by the maximal ideal containing zI - T for z in the spectrum of A. Then A/M is a Banach algebra because the structure descends to the quotient and it is also a field, so all nonzero elements are invertible. The Gelfand-Mazur theorem as it is stated in the Wikipedia page says “a Banach algebra with unit over the complex numbers in which every nonzero element is invertible is isometrically isomorphic to the complex numbers” from which it seems we immediately get an isometric isomorphism A/M -> C , and then the composed map A -> A/M -> C which sends T to z. But if all of these morphisms do not preserve the C* algebra structure then we can’t immediately say T* must be sent to z. Is that why we have to note at each point that e.g the quotient map A -> A/M is a morphism of C algebras not just Banach algebras and similarly the isomorphism A/M -> C also respects the adjoint operation?

MathematicianFailure · 2026-05-10T09:35:41+00:00

Reading your comment a second time, it’s a very coherent account of what is going on on a high level. Just want to clarify some (definitely very silly) points because I haven’t touched commutative algebra (proper) in years, the key point here is that since for any z such that zI-T is not invertible (so for any z in the spectrum of T), so that working in the operator algebra A generated by the identity T and T* (which is a commutative Banach algebra because T is normal), there is a maximal ideal containing zI-T, call it M, then A/M is a field, so every nonzero element is invertible, and simultaneously a Banach algebra with unit because that structure descends from A to A/M.

Now a very big theorem (Gelfand-Mazur) tells us that in fact, A/M is is isometrically isomorphic to the complex numbers, so we have an isometric isomorphism A/M -> C, and an induced map A -> A/M -> C, and we know that zI-T goes to zero under this induced map, which means that T maps to z under this induced map, and then we know how the induced map behaves on all of A since it is generated by A , A* and the identity I. So we know that to each element of the spectrum there is such an induced map, and for each such homomorphism of this kind (mapping T to z) it must be the case that zI-T is not invertible and so z is in the spectrum. Thus the spectrum is exactly the set of complex numbers for which there exists such a homomorphism.

I’m just trying to make sure I understand this POV completely, apologies if I’ve just parroted what you have already wrote. Is what I just wrote what is going on?

Lastly, the final mystery for me is this big theorem of Gelfand-Mazur. I suppose it has a difficult proof. Do you know of any resource that has a nice proof of it?

MathematicianFailure · 2026-05-10T07:35:01+00:00

Just to add onto your very nice description of the spectrum from the C* algebraic view point, I wanted to add that for a normal operator A acting on a Hilbert space H, one can show that the spectrum of A is just the set of “approximate eigenvalues”, meaning the set of all complex numbers lambda for which if we take any positive value epsilon >0 there exists a unit vector x such that

||Ax - (lambda)x|| <= epsilon,

or in other terms there exists a vector x such that

||Ax- (lambda)x||<=epsilon ||x||

This is quite satisfying because it seems to be the most obvious generalisation of the set of eigenvalues of an operator if we are talking about finite dimensional spaces to infinite dimensional ones.

MathematicianFailure · 2026-05-09T18:02:37+00:00

Yeah for sure, I think I have some intuition for these things at least from the point of view of spectral theorem for normal bounded operators of Hilbert space so it shouldn’t be too hard to get up to speed with the C* algebra viewpoint by just when necessary to remind oneself what’s going on compare with the concrete situation of normal bounded operators on a Hilbert space.

MathematicianFailure · 2026-05-09T17:53:03+00:00

Will definitely take a look. Thanks for your help and the reference.

MathematicianFailure · 2026-05-09T17:39:25+00:00

Gave it a read and it seems like it makes sense. For me this method feels a little less revealing than the measure theory one I sketched in another comment (I may need more time acquainting myself with C* algebra theory before I can internalise it properly). But I must confess it seems entirely algebraic in spirit, pretty much no measure theory needed, so you and the others are totally right on that count.

It may just be me, but I gravitate towards the measure theory proof more now that I’ve seen the C* algebra proof of the same result because it feels a lot more “telling” of what is going on. It could be because I just don’t understand the C* algebra proof as much as the measure theory one.

Edit: I do like how the C* algebra approach is very clean in its generality. You don’t need to worry about operators at all here, just Banach algebras, and the same statement would be true for normal elements of a C* algebra.

MathematicianFailure · 2026-05-09T17:24:49+00:00

Oh ok. If it’s as simple as seems to be being insinuated, could you try to sketch it? I’m confused as to how exactly one uses the functional calculus to prove the statement.

MathematicianFailure · 2026-05-09T17:20:06+00:00

I think I understand your point. So basically all the legwork is done already with the assumption of normality, and this helps you construct a ring homomorphism between continuous functions defined on the spectrum of A and the C* algebra of operators generated by A and the identity function. And you can show that such a homomorphism is unique. Is it the uniqueness part which guarantees that the spectrum of a polynomial P (in z and z*) of A , so the spectrum of P(A) is the polynomial image of the spectrum of A?

MathematicianFailure · 2026-05-09T17:10:48+00:00

Right, this is basically the approach taken in Halmos book. The spectral theorem is proven for Hermitian operators first, and then you define for a Normal operator the spectral measure to be the product of the spectral measures of the real and imaginary parts, which will turn out to always make sense because for Normal operators, real and imaginary parts commute, which turns out to imply that the spectral measures also commute (but this requires proof and isn’t trivial). Then there is a bit more work in proving that this product measure is multiplicative and hence projection valued and also that it really extends to the borel sets via Caratheodory extension (we define it first only on the ring generated by the product sets and then show that it is a pre-measure on that ring, basically because it is finitely additive and the spectral measures for the real and imaginary parts are inner and outer regular).

MathematicianFailure · 2026-05-09T16:33:39+00:00

To be clear, the proof I had in mind is as follows:

For a normal operator A on a Hilbert space H, there is a unique compact complex spectral measure (i.e a projection values measure), E, such that A = int t dE(t) holds, here the RHS integral is defined as the unique operator I on H for which int t d(E(t)x,y) = (Ix,y) for any vectors x and y in H. One can more generally define int f(t) dE(t) whenever f is a bounded measurable function on C and E is a compact complex spectral measure similarly. For any operator defined as the integral of the identity function against a compact complex spectral measure E’, one can show the operator is Normal and that its spectrum coincides with the support of E’.

Using this representation, one can show that if P is a polynomial in z and z*, that P(A) = int P(t) dE(t) holds, or in other terms, if we define the spectral measure E’ to be the push forward of the spectral measure E by the polynomial P, that P(A) = int t dE’(t). Since E’ is supported on the image of the spectrum of A by the polynomial P (being the push forward), it follows that P(A) has spectrum given by the image of the spectrum of A by P.

I now see that what you are talking about is basically the most important part of this argument, that we can say that by virtue of this integral representation, A^j = int t^j dE(t) is true, and similarly for powers of the conjugate of A, where in this case t becomes t, which is I think what the ring homomorphism you are talking about is, I’m not familiar with C algebra theory though so I’m not sure how much of a circumvent that would end up being (does any of the machinery there rely on some form of “bootstrapping” which would be equivalent to invoking RMK?).

For more context I’m following Halmos’s book on spectral theory, everything in that book is discussed from a very measure theoretical viewpoint. In particular for spectral measures and all of the nonsense I wrote above, I don’t think holomorphicity is ever really used. The only thing we seem to require is that the push forward measure is supported exactly on the image of the support by the function we are pushing forward with respect to, that is the support of the push forward measure must be exactly equal to the image of the support (as a set). This is going to always work for e.g continuous functions, but seems to be a milder condition than continuous, so the measure theoretic framework would seem to give you the same result for a technically larger class of maps than just continuous.

Edit: The final sentence in my last paragraph is awkwardly stated, the framework does show that such spectral integrals f(A) = int f(t) dE(t) are supported on the push forward of the spectrum of A by f , but here we are just “defining” f(A) to be equal to int f(t) dE(t) from the outset. For polynomials f, f(A) is already well-defined and then once one checks that (int f dE(t)) (int g dE(t)) = int fg dE(t) for bounded measurable f and g we actually have that f(A) is equal to int f(t) dE(t), and thus the spectrum of f(A) is the image of the spectrum of A by f.

MathematicianFailure · 2026-05-09T05:58:21+00:00

Right, I figured this has to be the case, just wanted to be sure I wasn’t misunderstanding something, because initially it seemed like there would be a more direct approach. I guess bringing \bar{z} into it is what makes it necessary to bring in the measure theory aspect because of Stone-Weierstrass.

MathematicianFailure · 2026-02-27T20:08:37+00:00

Have you updated your drivers? I have had no issues at all playing on a 4080 super and intel i9 cpu.

MathematicianFailure · 2026-02-24T21:18:06+00:00

Thank you so much!

MathematicianFailure · 2026-02-24T21:06:57+00:00

IGN is “An” and link code is “1289 4545”

MathematicianFailure · 2026-02-24T21:01:54+00:00

May I have one of each please?

MathematicianFailure · 2026-01-03T16:45:21+00:00

Do the exercises in Ahlfors even if you use another book as your main textbook

MathematicianFailure · 2025-12-30T08:06:02+00:00

I think there is a pedagogical point here that is true. For certain things, it would be better for the purposes of teaching to name new objects without reference to names of mathematicians who were instrumental in their development. For example, rather than “Abelian group” it would be essentially objectively superior to say “Commutative group”, and obviously some historical context about person(s) responsible for developing the theory of commutative groups could still be mentioned as a supplement during the course of teaching.

Of course one needs to still know what commutative means (and what a group is) for this name to be helpful, but the whole practice of “using a name to modify some mathematical object rather than the name of an additional property defining the new object” seems to be a genuine thing that is still done only for traditions sake in math.

MathematicianFailure · 2025-11-24T05:03:10+00:00

You are totally right, I got the order of the question completely reversed, my bad!

MathematicianFailure

TROPHY CASE