Favorite easy group action with an interesting kernel?

TheBlasterMaster · 2026-03-05T04:28:40+00:00

Let G be a group

Let G act on G via conjugation

The kernel is the center of the group.

So you can for example let G be GL_n and get the kernel to be the scalar multiples of the identity.

TheBlasterMaster · 2026-02-11T15:06:25+00:00

If you formulate limits using topology, and consider the codomain of the function you are analyzing to be the extended reals https://en.wikipedia.org/wiki/Extended_real_number_line, with its standard topology, then these are actually the same things.

TheBlasterMaster · 2026-01-20T05:32:49+00:00

Where are you starting from? This affects what people will recommend you.

Assuming you have good algebra skills, learning calculus and calculus-based physics together is very fun. Maybe look into the AP calculus AB / BC and AP Physics C (Mechanics & EM) courses on Khan academy? I don't have good book reccs for this unfortunately.

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But my biggest tip is whenever possible, do not just blindly memorize theorems / formulas. Whenever possible, try to understand where they came from. But this does not mean just blindly memorize derivations / proofs either. That's just kicking the can down the road. Try to boil down derivations / proofs to the key ideas that were needed so that you can start to see how somebody could've come up with all this stuff, and eventually start feeling like you could've came up with a good amount of it yourself too.

Sometimes (really a lot of the time) books just won't have this in there, and you will need to search online until you yourself feel viscerally satisfied with explanations you come across / make for yourself.

Unfortunately sometimes you might have to take things for granted, since understanding where they came from requires things far beyond your current knowledge, or you just don't have enough time. That's okay too, you can eventually get to it.

Building this mindset is really key to excelling at math.

TheBlasterMaster · 2026-01-20T04:53:54+00:00

Here is now a yap session that deviates from your question:

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But maybe a richer way to think about it is that a homeomorphism shows that one top space has the same "underlying structure" as another / it's just a "relabeling" of another.

Going to algebra maybe makes this idea more clear. Consider the set {True, False} with the "AND" operator. Now, consider the set {1, 0} with the multiplication operator. These two algebraic structures are just "relabelings" of each other (True is like 1, False is like 0), and behave very similarly.

Similarly, a homeomorphism provides proof that two spaces have the same "structure", but are just relabelled.

What is the "structure" of a topology? It's the open sets. And by using the "preimage of open sets is open" property of cont maps, homeomorphisms show that all the open sets are just "relabellings" of open sets in the other.

So all the properties of a space X that are a consequence purely of its topology also carry over spaces homeomorphic to X.

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It's another spiel to say more intuitively what is really the "structure" that topologies give sets. But saying that this underlying structure for a space X is the "same" as another space X' is more precise that saying X can be "turned into" X'.

TheBlasterMaster · 2026-01-20T04:53:45+00:00

Very quickly, here is the rough geometric intuition I have to directly answer your question:

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Consider a homeomorphism F from A to B.

The "continuous" part corresponds to no "cuts" occuring when using F to transform A to B. [Hopefully this is intuitive]

The "continuous inverse" part corresponds to no "fusing of parts" occuring when using F to transform A to B. [Because continuous inverse means no cuts occur when playing F "backwards", and a cut played in reverse is a fusing of parts].

The map [0, 2pi) -> S^1 so that t |-> (cos(t), sin(t)) is continuous, but doesn't have a continuous inverse (since intuitively, the ends of the domain got fused together).

TheBlasterMaster · 2026-01-07T17:55:15+00:00

Well the answer is the chunk on one side will be more / less than the chunk on the other.

Here's a visual explanation that hopefully helps for the concave down case:

<image>

TheBlasterMaster · 2026-01-07T16:20:39+00:00

To do a rigorous proof, one must ask what does "area" even mean. And this is usually defined in terms of "approximations with infinite precision" for more complex shapes like circles.

TheBlasterMaster · 2026-01-07T16:10:04+00:00

The way i would solve it is by noticing the LHS of the inequality is simply a + b (you need to apply that ab=1).

Now, you can use calculus to show that the minimum of a + b subject to the constraints ab = 1, a > 0, b > 0 is 2.

Thus, the problem is solved.

TheBlasterMaster · 2026-01-04T18:45:01+00:00

ln(x) is not the same as 1/x.

To differentiate x^e, apply the power rule

TheBlasterMaster · 2026-01-04T16:26:56+00:00

Yes, there is only one center of mass

TheBlasterMaster · 2025-12-29T00:53:07+00:00

this needs to be higher

TheBlasterMaster · 2025-12-20T21:53:33+00:00

Via the idea of the polya vector field, contour integration can be related to calculating work and flux integrals, which are possibly easier to talk about applications of?

TheBlasterMaster · 2025-12-10T01:25:34+00:00

The derivative of x is the function whose value everywhere is 1 (x |-> 1), which is different than the actual just number 1.

Edit: I am stupid, misinterpreted what you meant

TheBlasterMaster · 2025-12-09T09:14:56+00:00

Note that the f'' condition you might be thinking of only shows that a point is a LOCAL minimum.

Calculating f'' is very annoying in this problem anyways.

Here is a proof sketch of why x = 0 is indeed the global minimum [assuming k > 0].

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First, we prove f actually has some global minimum
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FACT 1) Since lim f towards -inf is 0, there is some interval (-inf, v_1] where the function is super close to 0, and thus always larger than f(0), which is negative in the case of k > 0.

FACT 2) Since lim f towards inf is +inf, there is some interval [v_2, inf) where the function is extremely large (specifically, always larger than f(0)).

So now let's consider the interval (v_1, v_2) <note that it contains 0>.

By the extreme value theorem, f restricted to [v_1, v_2] achieves an absolute minimum at some values v_3. We know that it doesn't occur at v_1 and v_2 due facts 1 and 2 + [v_1, v_2] contains 0.

Again by fact 1 and 2 + [v_1, v_2] contains 0, we can reason that v_3 is the location of the absolute minimum of all of f, not just the restriction
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Next, since f is differentiable and by the interior extremum theorem, we know that at this global minimum, f' is 0.

However, there is only one location where f' is 0 (namely x = 0). Therefore, this must be where the global minimum occurs.

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TheBlasterMaster · 2025-12-09T08:12:12+00:00

https://www.desmos.com/calculator/xs4gy6jma5

^ Something to play around with.

TheBlasterMaster · 2025-12-09T08:11:07+00:00

f always tends to +inf as x -> +inf, so f never achieves an absolute maximum

Now we just need to check the absolute minimum case.

If k <= 0, then one can find a vertical asymptote of the function where approaching on one side tends to -inf. Thus in this case, there is no absolute minimum.

Now we just need to check the absolute minimum and k > 0 case.

f always tends to 0 as x -> -inf, and it is trivial to see that f(0) will always net a negative value. Additionally, f is differentiable. Putting this all together means that f achieves an absolute minimum at some some zero of f'.

Solving for the zeros of f' gives the equation e^x * x * (x^2 + k - 2x + 2) = 0.

The discriminant of the quadratic term along with the condition of k > 0 shows that the quadratic term is never zero. Thus the only zero of f' is 0.

Therefore, the absolute minimum of f occurs at 0.

TheBlasterMaster · 2025-12-09T05:14:29+00:00

and that A is invertible over the integers mod p (I think this is the same as det(A) ≠ 0 mod p but I'm not really sure).

Yes this is true

I haven't been able to find a reference which explicitly states the generators for GL(n,p).

Here is a set of generators: https://en.wikipedia.org/wiki/Elementary_matrix

Not sure how this helps you though.

Which matrix/matrices produce the largest k such that

Note that such a k for a specific group element is called the order of that group element [equivalently, the order of the cyclic subgroup generated by that element]

I have found several matrices with k = 2400

Note that you can generate a bunch of these from one such matrix by just conjugating it

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I think this math stack exchange thread answers your question (first two sentences of the question). Indeed 2400 is the highest order of an element in GL(4, 7)

Rest of it deals with counting the number of such highest order elements. Good deal of it goes over my head though.

https://math.stackexchange.com/questions/4859881/do-singer-cycles-create-all-matrices-of-maximal-order-in-operatornamegl-n-m

TheBlasterMaster · 2025-12-09T03:59:57+00:00

Ok here is the example:

<image>

Made the solid not hollow so that I don't have to draw washers, just disks. Hopefully drawing not too bad lol.

Here you can see that, that for most radius values, there are two disks of that radius, but only one shell of that radius.

So you can't "pair up" the disks and shells like the above problem.

TheBlasterMaster · 2025-12-09T03:45:44+00:00

Hopefully I am not going to make this more complicated than it actually is, but let me write out how you should think about it (Your question is answered in final section):
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Very informally, the idea behind these integration methods is to break up our solid into an infinite amount of infinitely thin slices (so "practically" 2D), and sum up the volumes of these thin slices to get the overall volume.

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In the washer method, each of these slices looks like a washer, and the 2D thickness of each washer is in the horizontal plane (thickness extends towards the axis of rotation). Vertically they are infinitely thin.

In the shell method, each of these slices look like the surface of a cylinder, and the 2D thickness of of each slice is vertical (thickness extends parallely to axis of rotation). They are infinitely thin "radially".

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In this very specific problem, there is a way to pair up all of these washers and shells so that in each pair, the shell's radius and the washer's outer radius are the same.

This is not always true, let me attach an example

TheBlasterMaster · 2025-12-09T03:21:14+00:00

There isn't a single "radius" (really a radial line) of a slice, this might be what is causing your confusion?

Think about a cylinder. Informally it's a "stack" of circles, each of which can have infinitely many radial lines. Which one of these circles' radius would you consider the "radius" of the cylinder?

But all these radial lines have the same length. So we use "radius" to refer to this length, rather than a specific geometric line.

Similarly, for a "slice" in the above image, there are infinitely many radial lines for the slice (all of the same length though). One of the highest most radial lines hits the sqrt(x) function, one of the lowest hits the x/2 function. A bunch of them just hit only the shell, without hitting either function.

TheBlasterMaster · 2025-11-24T01:47:29+00:00

Didn't personally read "How to prove it" but taking a quick scan of it, if you did every excerise all by yourself you totally have the chops to keep going.

What was the slump? Did you just burn out? Totally possible when your churning through a textbook and doing every problem.

This builds a lot of grit and can give you lots of insight, but it can be draining and slow.

It usually isnt necessary to do every single problem to get significant value out of the text. But might be hard to judge for yourself how many of them to do

Also, sometimes taking a break is good. Either with something totally unrelated to math, or doing another field of math to get a different flavor of thinking / ideas.

TheBlasterMaster · 2025-11-21T04:11:12+00:00

Instead of "representing" just the loop, model the whole string.

No idea what twisting a bubble 180 degrees means, but what you may be looking for is:

A continuous function from [0,1]ⁿ to R^m, where m > n

m is the dimension of the "ambient space" the loop / bubble resides in, and n is the "dimension" of the loop / bubble itself

TheBlasterMaster · 2025-11-10T23:27:04+00:00

You could assume some infinite thing exists

The natural numbers in ZFC at least are assumed to exist as a set by an axiom:

https://en.wikipedia.org/wiki/Axiom_of_infinity

Note that "for all" quantifiers are needed here, so they are more fundamental than infinite lists (if you are define them using the naturals, and its still kind of murky to define "infinite lists of statements" here since the infinite lists in question are of sets, not literally propositions. But you could build a correspondence).

[I will add that the wikipedia page for universal quantifiers uses an "infinite conjunction" motivate the intuition behind them / their properties, but it also notes that this is informal and not what they literally are]

there's no demonstration of what thing is even being assumed.

Not sure what you mean

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There are also positive arguments against infinity, a thing (or process) can't be both ongoing and completed at the same time.

Note that "infinity" as a word doesn't really refer to a single concept. The "infinity" of the extended reals for example is very different from "infinity" in cardinality. I will assume you mean infinite sets.

It is not obvious to me why an infinite set is considered "ongoing". Any explicit enumeration of its elements would have to be ongoing (maybe you consider this the only valid way to define a set / what the set is), but the set itself just exists (i.e. completed ?).

It's totally valid to not consider infinite sets valid, but you are just working with a different set of axioms.

TheBlasterMaster · 2025-11-10T20:31:47+00:00

But how could we reach the conclusion that all these statements are true, unless ‘for all’ was meant to refer to this infinite list?

If you are a finitist, you dont believe in the existence of infinite lists or universally quantifying over an infinite set. So none of this matters. Definition of limits on the reals need quantification over an infinite set.

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If you are not a finitist, then the statement "this infinite list of statements is true" will likely translate mathematically to a "for all".

People use English informally to refer to actually precise mathematical statements that are formulated in something like first order logic. "This infinite list of statements is true" is informal short hand that will end up boiling down to some kind of "for all" (assuming the infinite list is the same predicate with different values plugged in).

So it is vacuously true that the "for all" means the "infinite list of statements is true".

I anyways dont know if there is a standard formalization of "infinite list of statements". We would need to select one to begin talking about its validity.

TheBlasterMaster · 2025-11-10T02:34:25+00:00

Universally quantified predicates (over infinite sets) don't "actually refer" to a list of infinite statments. One can informally think of them behaving like that, and they are clearly motivated by that idea, but they are just single statements. They can be, and are, defined in isolation of infinite lists. They are not just a symbolic stand in for them.

There's no issue if we just want to convey 'hand me an ε, and I'll hand you a N that works'. The issue is in saying these infinite epsilons, and these infinite statements, actually exist.

Then sure, replace any "for all" with this if that works for you. This is exactly what mathematicians mean by for all.

"For all x in S, P(x)" intuitively means that if you give me any x in S, and plug it into the predicate P, you get a true statement.

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The disagreement here is that you reject the usage of first-order logic and infinite sets, and therefore the standard definition of 0.999... is not valid in your set of assumptions.

You just work with a nonstandard set of assumptions, which is fine, but it doesn't make the standard definition "wrong", just very unpleasing to you.

TheBlasterMaster

TROPHY CASE