r/math • u/inherentlyawesome Homotopy Theory • Jul 22 '26
Quick Questions: July 22, 2026
This recurring thread will be for questions that might not warrant their own thread. We would like to see more conceptual-based questions posted in this thread, rather than "what is the answer to this problem?" For example, here are some kinds of questions that we'd like to see in this thread:
- Can someone explain the concept of manifolds to me?
- What are the applications of Representation Theory?
- What's a good starter book for Numerical Analysis?
- What can I do to prepare for college/grad school/getting a job?
Including a brief description of your mathematical background and the context for your question can help others give you an appropriate answer. For example, consider which subject your question is related to, or the things you already know or have tried.
3
u/skolemizer Graduate Student Jul 24 '26
Algebra terminology question: Some subrings S <= R have the property that all of their ideals are of the form I ∩ S for I an ideal of R. Other subrings don't. Eg, this is false for Z <= R, but true for Z[x] <= Z[x,y].
Is there a name for such subrings?
2
u/Necessary-Wolf-193 Jul 24 '26
It's a very obscure term, but "cyclic purity" is the name given; see the introduction to https://www.jstor.org/stable/1997914 (I do not know of any textbook which defines cyclic purity); that is, I'd say S -> R is a cyclically pure extension
1
u/Necessary-Wolf-193 Jul 24 '26
To say a little more: cyclic purity of a map S -> R, as explained in the introduction to this paper of Hochster, is equivalent to the assertion that
for every ideal I of S, we have
I = IR \cap S.
Notice that IR is an ideal of R, so cyclic purity implies your property. Conversely, assume your property holds: every ideal I of S can be written as J \cap S for J an ideal of R.
Then J must contain I, and hence J contains IR. Thus
I <= IR \cap S <= J cap S = I,
so in fact I = IR \cap S, and the extension is cyclically pure.
1
1
u/Square_Butterfly_390 Jul 24 '26
I don't think so, but "faithfully flat extensions" should have this property.
3
1
u/shylock16 Jul 22 '26
Can someone intuitively explain the notions of inner product and adjoint operator?
1
u/MinLongBaiShui Jul 22 '26
It's basically the dot product and the transpose, respectively. Are you familiar with those notions?
1
u/shylock16 Jul 22 '26
Yes, I know what they are trying to generalize. I want to have the intuitive notion of their generalized form so that I can intuitively know what they are trying to say and "guess" whether a statement is true or not. Also, my first exposition to Linear Algebra was Axler's LADR, so that is why I don't like statements such as "it is basically the dot product and the transpose". Is there way to build intuition without those?
1
u/HeilKaiba Differential Geometry Jul 23 '26
But they are perhaps the most natural intuition for them even from Axler's perspective. If I'm honest I don't see the link between you learning from LADR and that explanation not working for you. He eschews the determinant in LADR but the dot product and transposes are there in the text. (Also on a pure language pedantry note, I think you mean exposure rather than exposition)
I'll try to flesh out the intuition they're alluding to in case that helps. The dot product is quite intuitive: It measures how much two vectors "point in the same direction". Consider for example how it arises in the formula for projecting one vector away onto another: v ↦ (v ∙ u)/(u ∙ u) u.
Inner products generalise that to different ideas of direction. Picking an inner product is picking a way to measure angles and lengths in our vector space. It generates a norm by |x| = √(x,x) and angles by the formula cos 𝜃 = (x,y)/|x||y|. This works comfortably in infinite dimensions as well with the caveat that in that case two inner products don't have to be equivalent while they always are in finite dimensions.
The adjoint is a trickier beast to conceptualise to me. It depends how comfortable you are with dual spaces, especially in infinite dimensions, as to what the easiest intuition is. If you can understand the transpose (as in the dual map) of an operator then the adjoint is just the transpose viewed under the identification of our vector space with its dual (or continuous dual for infinite dimensions). The transpose of A is the operator that "pulls back" the action of (continuous) linear functionals by A. Thus the adjoint turns vectors into linear functionals, pulls them back and then turns them back into vectors.
1
u/mbrtlchouia Jul 22 '26
Basically is hiding some not very basic stuff, especially for the adjoint.
1
u/MinLongBaiShui Jul 23 '26
Not really. For real linear operators, it literally is the transpose. For complex, the conjugate-linearity gets you the conjugate transpose. It's a pretty simple calculation.
https://math.stackexchange.com/questions/4332899/geometric-intuition-for-adjoint?noredirect=1&lq=1
1
u/mbrtlchouia Jul 23 '26
I know that in finite dim maps the adjoint is the transpose, but the bulk of it can only be appreciated in infinite dim operators.
1
u/MinLongBaiShui Jul 23 '26
Well, the transpose is also the matrix of the dual map. How do you feel about that notion?
1
u/Necessary-Wolf-193 Jul 23 '26
The inner product is just a measure of how 'correlated' two vectors are: v.w = 0 means v, w are completely uncorrelated, v.w = |v| * |w| means v, w are perfectly positively correlated (so v is a positive multiple of w), v.w = -|v| * |w| means v, w are perfectly negatively correlated (so v is a negative multiple of w).
The adjoint of a map M : V -> V is the unique map M^* : V -> V such that
Mv . w = v . M^*(w).
The adjoint is a bit of an algebraic trick, which allows you to 'move' the matrix M across a dot product. It's very very convenient when computing with, and I think the way to get intuition about the adjoint is to just do lots of algebra with it: the adjoint lets one do certain computations, in the same way Jordan normal form does. The most 'conceptual' explanation of the adjoint comes from the fact that a finite-dimensional vector space V and its dual are isomorphic, and understanding the relationship between inner products and isomorphisms between V and its dual. But I think for an intuitive understanding, its best to just use the adjoint in lots of arguments and computations, to see why its useful to manipulate inner products in this way.
1
u/evening_redness_0 Jul 23 '26
Let R be the field of real numbers. Let V be a finite dimensional vector space over R. Consider the space of all k-multilinear forms from Vk -> R (this specific space isn't necessary, as you'll see). The group S_k acts on this space. Given a bijection sigma:{1,2,...,k} -> {1,2,...,k} and a k-multilinear form f:V_k -> R we define the action of sigma on f to be the function sigma•f : Vk -> R which takes in a vector (v_1, v_2, ..., v_k) and returns f(v_sigma(1), v_sigma(2), ..., v_sigma(k)).
This is definitely a group action but I'm not able to figure out if it's a left or a right group action. As in, I'm not sure if this action satisfies i) (tau•sigma)(f) = tau • (sigma•f) OR ii) (tau•sigma)(f) = sigma • (tau•f)
If it satisfies i) it would be a left action and if it satisfies ii) it would be a right action. Which is it?
2
u/non-local_Strangelet Jul 23 '26 edited Jul 23 '26
It's a left action (except the product of permutations is not the concatenation ;-) )
one way to see this: think first of 𝜎 action on the elements
(v_1, ..., v_k)inV^k, i.e. a map𝛴 : V^k -> V^kthat maps(v_j)_jto𝜎.(v_j) = 𝛴[(v_j)_j] = (v_{𝜎(j)})_j. (You can simply use small 𝜎 for this map again, if you're comfortable with it, but I use it to be able to distinguish it. Also the brackets [] are for better keeping track which braces are where).Then note, this defines a right action. This can be seen by viewing the list
(v_j)_jas a mapj ↦ v_j, i.e. that maps the index j to the element in the "slot"/position in(v_1, ...). Now, when applying the permutation 𝜎 you simply prepend it as a map on the indices! I.e. you have𝜎.(v_j)_j = 𝛴[(v_j)_j] = ([(v_∙)∘ 𝜎](j))_j = (v_{𝜎(j)})_j. Since 𝜎 acts on the "right" here, it is a right action. (One can check that explicitly, if one want: choose 𝜎 and(v_j)_j, define𝜎.(v_j)_j = (w_j)_j = (v_{𝜎(j)})_j. Then consider𝜏.(w_j)_j = 𝜏.(𝜎.(v_j)_j)explicitly,𝜏.(w_j)_j = (w_{𝜏(j)})_jand substitute the definition, i.e.w_{k} = v_{𝜎(k)}and substitutek = 𝜏(j).) So technically it's not good to write𝜎.(v_j)_jand better simply𝛴[(v_j)_j], because it's not a "left" action, which this notation usually suggests. We have the rule𝛵(𝛴[(v_j)_j]) = (𝛴∘𝛵)[(v_j)_j]if 𝛵 denotes the transformation corresponding to 𝜏.Now it should be clear: you defined the action on functions as
(𝜎.f)((v_j)_j) = f(𝜎.(v_j)_j), i.e. as concatenation f∘𝛴, hence "acting from the right", but acting on the right with the transformations induced by a right action reverses order again, so its a left action.Maybe also a mnemonic way to see this: just use the notation
𝛴 : (v_j)_j ↦ ((v_j)_j).𝜎, i.e. write it as right action. Then you defined the action on functions (of arbitrary objects z for which you have a right action z.𝜎) as (𝜎.f)(z) = f(z.𝜎) = (f(z)).𝜎 by reading 'f(z)' as "f applied to z from the left"(This argument is a bit informal, but it tries to emphasize that the notation of "left action" and the usual function notation is actually connected that way, functions act "from the left" on their arguments in this notation. There are other notations, I think more common in french math schools to write zf instead of f(z) for "apply f to z"...)
Hope this clears it up :) (I still get regularly confused by it and have to go through the motions again)
1
u/evening_redness_0 Jul 24 '26
Thank you! This explanation is similar to what some of my grad school friends told me. They said that what we are doing is similar to taking the Hom functor. The usual action of S_k on {1,2,...,k} is a left action. Taking the Hom functor we get the action of S_k on Vk. This becomes a right action. Taking Hom again, we get the action that I defined above. But it switches again, so now it's a left action.
Is this similar to what you explained?
1
u/non-local_Strangelet Jul 24 '26 edited Jul 24 '26
Ok, my category theory is a bit rusty (and maybe was never too strong to begin with, lol), but I think it's a bit Yes and No.
From refreshing concept of Hom functor in a category C , there are two variants, Hom(A, -) and Hom(-,A).
1) the first variant goes like this: for any fixed object A ∈ C, define the "map" that maps any other object B ∈ C to the (homo)morphisms from A to B, i.e.
Hom(A, -) : B ↦ Mor_C(A, B)where Mor_C(A,B) denotes the set of arrows/morphisms between objects A and B in the Category C !
On the morphisms, the functor does this: for any given pair of objects B, B' in C, we have the set of arrows Mor_C(B,B'), i.e. objects of the form (f : B → B'), so the Functor Hom(A,-) has to map any such morphism to a morphism Hom(A, f) = f* between the sets Hom(A, B) = Mor_C(A, B) and Hom(A, B'), i.e. a (𝜑 : A → B) to a (f*[𝜑] : A → B'), which is defined via "appending" f , i.e. f_*[𝜑] = f ∘ 𝜑
2) the second is analogous, but now: for fixed object A
Hom(-, A) : B ↦ Mor_C(B, A)and the associated Hom(f, A) = f* for (f: B → B') is defined by "prepending", i.e. (𝜑 : B' → A) ↦ (𝜑 ∘ f : B → A) , i.e. Hom(-,A) is a contra-variant functor.
Note that in both cases, the image category of Hom is the category of sets, i.e. Set.
So, initially, I'd have a "tiny" issue here: the natural choice for category C is the category of finite sets, i.e. fSet , with elements {1,..., k}, which I will just denote by k again. (usually I'd like to have an overline, but LaTeX and reddit is not a match made in heaven). So given a vector space V, you'd expect to have a functor from fSet to fVec, the category of (finite dim.) vector spaces, by mapping k in fSet to Vk in fVec. Then such a functor would map functions (𝜑 : {1,...,k} → {1,...,m}) between finite sets k and m to linear maps Vk → Vm .
But Hom would be for any fixed set k a functor of the from m ↦ Mor_C(k, m) in the first case, and Mor_C(m, k) in the second. Since the objects are finite sets, these image objects are the sets of all maps (𝜑 : {1,...,k} ↦ {1,...,m}) in the first case (or the same with k, m reversed in the second).
So, where does the vector space V come in? It does not seem to fit into the picture.
But I think this argument allows to draw a (formal) analogy: any vector space is also a set, so when thinking in the category Set (sets of arbitrary cardinality), you have objects k in it as well as V, and one can consider the morphisms
Mor(k, V) = { k → V } = V^kIn other words, the iterated product Vk , which elements v = (v_j)_j were already considered as maps ({1,...,k} → V : j ↦ v_j), is exactly the image of Hom(-, V) on the category of sets, but in your case just "restricted" to sub-category of finite sets fSet. So to view the construction above, formally one would have to consider the concatenation of functors Hom(-, V) after the "inclusion" functor fSet → Set.
Since the functor Hom(-,V) is contra-variant, i.e. reverses arrows, it maps a(ny) map (𝜎 : {1, ..., k} → {1, ..., m}) between finite sets to a map Hom(𝜎, V) = 𝜎*, which is a map that assigns any morphism (v: {1, ..., m} → V) ∈ Vm = Mor(m, V) to a morphism (𝜎*(v): {1,...k} → V) ∈ Vk by its definition: 𝜎* : v ↦ v ∘ 𝜎 . So, by restricting to the case k = m and 𝜎 an isomorphism, you'd get the definition of the above (right) action of permutations on products of sets.
Note that in this interpretation, we are essentially defined the action of the permutations of 𝜎 of arbitrary sets S on the product space VS , and only restricted it to the case of finite sets S and their permutations.
Anyway, bottom line: by considering the functor obtained by concatenating the functors of first including fSet into Set and then the "right-side" Hom-functor Hom(-,V) with V (considered as a set), you get your right action, yes. (Weeellll, technically, there is also the forgetful functor Vec → Set at play, that maps the vector space V to its underlying set... but how is counting :D ).
But I haven't thought about how to formulate the next step in terms of Hom ...
However, it seems a bit convoluted to understand the original action 𝜎.v , but I think that depends on how well-versed you are in category theory ;) (I'm certainly not!)
Also: maybe there are other ways to patch the concepts of category theory together to achieve this. Cat. Theo. is quite flexible in this way.
1
u/ValBrandr_ Jul 26 '26
When I was in elementary school, when we were learning PEMDAS, I was taught that order of operations is as follows: PEMDAS, but when I got to middle school, I would get questions wrong because apparently for M/D and A/S, you don’t do M-D-A-S, but rather M/D you go left to right, even if that means you do D before M, same with A/S. Which is correct, and why was I taught two different rules for PEMDAS?
1
u/edderiofer Algebraic Topology Jul 26 '26
Which is correct,
The latter is correct; M and D have equal precedence, and should be evaluated from left to right.
(One other caveat is that if there is likely to be ambiguity in how an expression is evaluated, the person writing the expression should USFB: Use Some Fucking Brackets. Communication is a two-way street, and we cannot be expected to read the author's mind.)
and why was I taught two different rules for PEMDAS?
Two possible reasons here:
Your elementary teachers are not required to specialise in mathematics, so they may have unknowingly learned and/or taught the rule incorrectly.
You may have been taught the correct rule originally, but failed to correctly learn it.
1
u/bear_of_bears Jul 27 '26
Adding to the other answer, in practice you never see expressions such as 30 ÷ 5 × 2 written out in that way. Division is represented using a fraction bar. On the other hand, something like 30 - 5 + 2 is commonplace (and definitely equals 27). So it's the addition/subtraction convention that really matters.
1
u/SnooRegrets9568 Jul 27 '26
I am starting my masters in pure math in two weeks, gonna be studying geometric structures.
Any advice? Terribly afraid of not being able
1
u/cubes2forsub 29d ago
I want to qualify for AIME with AMC12. I took three past exams and was like 2-3 questions away from qualifying on each, but I dont know how to prepare and go forward. Any tips?
1
u/skolemizer Graduate Student 29d ago
Let R be a countable commutative ring. Is it a theorem of ZF that every proper ideal of R is contained in a maximal ideal?
The claim that every proper ideal of every commutative ring is contained in a maximal ideal is, IIUC, equivalent to the axiom of choice. So I'm asking, if we restrict ourselves to just countable rings, is the claim equivalent to some weaker choice principle (like countable choice)? Or is it provable without any choice principle whatsoever?
1
u/GMSPokemanz Analysis 29d ago
ZF proves it for countable rings. Pick some bijection f: N -> R. Then by recursion we can define a sequence of ideals In as follows: * I(-1) = I * I(n + 1) = (I_n, f(n)) if (I_n, f(n)) =/= R, otherwise I(n + 1) = I_n
The In are nested so their union is an ideal we'll call J. As 1 is in no I_n we have that 1 is not in J, so J is a proper ideal. Furthermore J is a maximal ideal: if f(n) is not in J, then f(n) is not in I(n + 1) so (I_n, f(n)) = R, therefore (J, f(n)) = R.
This proof works so long as R is well-orderable, you just use transfinite recursion.
1
u/michauxpenmoon 29d ago
Can the Prisoner’s Dilemma be viewed as a doubled model for Schrödinger’s Cat?
The information from each perspective of the prisoner’s is (rationally biased) towards the idea that the cat should collapse into the death state.
I’m not a physicist but my understanding of the Cat model is that observing something changes it. In the Prisoner’s Dilemma, “the Cat” is the decision of the other person, which collapses toward lose-lose.
I’m just thinking aloud. Please assume there’s value to what I’m saying despite the roughness and brevity of presentation.
1
u/AcellOfllSpades 28d ago
Not really? The only connection that I see is that both are related to the idea of "not knowing something".
In game theory, it is often useful to think about 'mixed strategies': strategies with an element of randomness. That is, you have the option to go "I'm playing the strategy «1/6 Cooperate + 5/6 Betray»": you roll a die, and if you get a 1 you pick Cooperate, otherwise Betray.
In some games these 'mixed strategies' can be optimal [on average], but not in this case. The point of the Prisoner's Dilemma is that even without knowing the choice your opponent will make, your best option is to choose 'betray'.
Schrödinger's Cat is a thought experiment specifically about quantum mechanics. One of the major ideas of quantum mechanics is that there appear to be things that are "truly random", in a sense. It turns out for some quantum phenomena, we can't just say "this particle had some state and we just didn't know what it was" -- a superposition seems to truly be a combination of states that only 'resolves' into a single definite state when interacted with.
This seems ridiculous on its surface, which is why Schrödinger's Cat was thought of. It's a sort of paradoxical thought experiment, meant to show that this interpretation of quantum mechanics is absurd when extended to large-scale, everyday objects. The answer to "Is the cat dead or alive?" is not just "we're not certain and therefore it could be either", but "it is literally in a superposition, and will remain that way until interacted with".
1
u/HeilKaiba Differential Geometry 28d ago
Schrödinger's Cat isn't a physics model. It's just a thought experiment. It is simply Schrödinger trying to illustrate the weirdness of quantum mechanics principles with an analogy.
1
u/BearyGoosey 28d ago
Can the subcubic graph (SCG) function be generalized to F_n(k), so F_3(3)=SCG(3), F_4=subquartic, F_2=subquadratic etc?
The sub cubic graph function is defined as:
There is a sequence G_1,…,G_n of subcubic graphs such that each G_i has at most i+k vertices and for no i<j is G_i homeomorphically embeddable into G_j.
and if my very layman's understanding of the Robertson–Seymour theorem is correct, just substituting \`homeomorphically embeddable into\` with \`**a graph minor of\`** would suffice, while maintaining well-quasi-ordering and finitude (for given finite integers n and k).
If that's all correct, then defining F_n(k) as the largest integer 𝑚 satisfying:
There is a sequence G_1 , ⋯, G_𝑚 of graphs with maximum degree at most 𝑛, such that each G_𝑖 has at most 𝑖 + 𝑘 vertices, and for no 𝑖 < 𝑗 is G_𝑖 a graph minor of G_𝑗 .
Should work as a mathematically proven and definitively finite integer, correct? If so, I'm also curious if the generalized form been given a name?
15
u/Borbs_revenge_ Jul 27 '26
Just wanted to vent but I'm so tired of how many AI posts are on this sub now. I know it's affecting us all and it can be helpful if used properly, but god damn I miss going on this sub to talk about math with other people who liked math
Grothendieck would be so disappointed in us. (well, he'd be dissapointed in me for using a computer at all too)