@Triskelion When they say $g$ induces an inner automorphism of $A_5$, they mean that there is some $a \in A_5$ so that for all $b \in A_5$, we have that $g b g^{-1} = a b a^{-1}$ --- the automorphism $\varphi_g$ is in fact inner as an automorphism of $A_5$. I suspect your confusion comes from the fact that $\varphi_g$ is defined in terms of conjugation by some element, so looks like what we would call an inner automorphism --- but that element doesn't lie in $A_5$.

Secondly, if your semidirect product $N \rtimes_\varphi H$ comes from an inner automorphism $\varphi$ --- meaning we have $f: H \to N$ so that $\varphi(h) n = f(h) n f(h)^{-1}$ --- then it's isomorphic to the direct product. Here is the formula for the isomorphism: we send $(n,h)$ in $N \rtimes_\varphi H$ to $(n f(h), h) \in N \times H$. You can check that this is a bijective homomorphism.

In your case, $H = C_2$, so the map $f: H \to N$ is just the choice of square-1 element $g \in A_5$.

Then the statement that $\varphi_g = \varphi_a$ up above means that the map $C_2 \to \text{Aut}(A_5)$ is inner, as in my previous comment (it comes from the map $f: C_2 \to A_5$ with $f(1) = a$ --- I suppose I'm assuming here that the element inducing the inner automorphism has order 2, which is not necessarily true. I'll think about this in a moment)

Which would imply the semidirect product is in fact a product.

Ah, and the point is that because $A_5$ has trivial center, the fact that $\varphi_{a^2} = \varphi_a^2 = 1$ means that $a^2$ comments with every other element of $A_5$, and so is the identity. That's why $a^2 = 1$, which is what we need for $f$ to be a homomorphism.

Sorry if my notation is idiosyncratic.

@Triskelion Yes, though these can be much more complicated than semidirect products. There's a particularly interesting nontrivial central extension $A_5 \to G \to C_2$ (in particular, $G$ is not $S_5$ --- not central --- nor $C_2 \times A_5$ --- that's the trivial extension)

Indeed; and something like the same argument should show that if two maps $H \to \text{Aut}(N)$ "differ by an inner automorphism" then the two semidirect products are isomorphic, where that isomorphism fixes $H$ as a set.

There's the sign homomorphism $s: S_5 \to C_2$ with kernel $A_5$ and any 2-cycle gives a section of that, right?

Those are the most commonly discussed kinds of extension because they're the easiest to classify. But there are many. Your comment that central + split => trivial extension is correct.

Sorry, I think my terminology is throwing you off.

By section I mean splitting. A map $f: C_2 \to S_5$ with $s(f(1)) = 1$.

By 2-cycle I mean transposition like (12).

$f$ is supposed to be a homomorphism!

So definitely not

I'm not sure why you think that the conjugating element should be independent of the input element of $H$ --- we're trying to produce a homomorphism $H \to \text{Aut}(N)$

That takes an element of $H$, and gives us an automorphism of $N$; so if we have an $h$ and an $n$, then it spits out a new element of $N$, $\varphi_h(n)$. That will depend on $h$

in such a way that $\varphi_h(\varphi_k(n)) = \varphi_{hk}(n)$

(which would be impossible if $\varphi_h$ was independent of $h$, unless $\varphi_h = 1$ for all $h$)

This is all notational trouble you're running into. An inner automorphism is a map of the form $\varphi_a: N \to N$, $\varphi_a(n) = ana^{-1}$.

That's an element of $\text{Aut}(N)$.

We are producing a homomorphism $\varphi: H \to \text{Aut}(N)$. That means for each element of $H$ we spit out an automorphism.

Given the homomorphism $f: H \to N$, we define $\varphi(h)$ to be the inner automorphism $\varphi_{f(h)}$.

Each one of these is an inner automorphism; conjugation by the element $f(h)$. But the automorphism we get depends on the element of $H$.

I've just checked that there are only three groups up to isomorphism with composition factors $C_2$ and $A_5$; they are $C_2 \times A_5$, $S_5$, and a group called "$2I$". I'm sure there's an elementary argument, but I didn't write one down

Ah yeah I can but I'll stop doing this and let you stop and think

Well if there didn't, then $\varphi_g$ is not an element of $\text{Inn}(A_5)$, so it's not a trivial semidirect product

If $A\to B$ and $B\to C$ are ring maps with the kernel of both finitely generated ideals, should the kernel of their composition $A\to C$ a finitely generated ideal in $A$?

man rip i did not know you can only ask 6 questions in 24 hours

Solved.

@Triskelion I don't follow. An element $g \in N$ induces an automorphism of $N$ by the formula $\varphi_g(n) = gng^{-1}$ --- that makes sense because $N$ is normal. But why should there be some $a \in N$ so that $gng^{-1} = ana^{-1}$ for all $n$? How would you produce it?

Sorry! The second sentence should begin "an element $g \in G$".

@Hokie: Some people definitely abuse their privileges.

@Triskelion Your question doesn't make sense to me?

@Triskelion This was your original question.

That means that the automorphism of $N$ given by $\varphi_g(n) = gng^{-1}$ is not in fact of the form $\varphi_a(n) = ana^{-1}$ for $a \in N$. There's no reason that this should be true a priori for some random $g \in G \setminus N$.

@TedShifrin is there a way I can ask 1 more question? I only have 1 more and my final is due at midnight

That means it's not inner. That means it's not of the form I was describing. That means that the semidirect product is not isomorphic to the direct product.

@Hokie: So you're putting your exam questions on MSE for people to solve them for you? Seriously. Go away.

That is serious cheating.

No. I'm describing a special kind of homomorphism $\varphi: H \to \text{Aut}(N)$.

Given a homomorphism $f: H \to N$, this gives you a homomorphism $\varphi_f: H \to \text{Aut}(N)$, called "conjugation by $f$", defined by $(\varphi_f(h))(n) = f(h) n f(h)^{-1}$.

These are relevant because such homomorphisms have $N \rtimes_{\varphi_f} H \cong N \times H$.

You were asking about a split extension $A_5 \to G \to C_2$; call the splitting $s$, so that $s(1)$ is an element of $G$ with $p(s(1)) = 1$. You separated this into two cases: 1) the case that $g = s(1)$ induces an automorphism of $A_5$ which is of the form $\varphi_a$, for some $a \in A_5$.

In that case your semidirect product is $A_5 \rtimes_{\varphi_f} C_2$, where $f(1) = a$. That is the relevance of the discussion of these homomorphisms $\varphi_f$.

This means that in that first case, your semidirect product is isomorphic to $A_5 \times C_2$.

In the second case, $g$ does not induce an automorphism of the form $\varphi_a$. This seems to be what you were just asking me about --- you were asking why it's possible that there's no such $a$. I think the question is why there should be such an $a$.

Yes. What does that have to do with case 2?

Do some people define the projection map of a vector bundle $\pi : E \to M$ without having the condition that $\pi$ must be subjective?

subjective? :D

I sure hope not.

It annoys me enough that Hatcher doesn't require covering maps to be surjective.

Oh lol I meant surjective

Do you have evidence than someone has committed this blasphemy?

@Perturbative: It must be surjective onto connected components or else miss them entirely. Not interesting. (Submersions are open maps, remember.)

@TedShifrin Milnor and Stasheff but they mention after they define it that the fibers are never empty so it's equivalent to surjectivity

@TedShifrin So whether his maps are covering maps is indeed subjective

@Alessandro: I have to concur.

Take my star @AlessandroCodenotti

And my axe @Perturbative

@Perturb: So we agree it's not an interesting question. Next.

@Triskelion Write the two elements of $C_2$ as $[0]$ and $[1]$ to be clear what they are --- then $[0]$ is the identity element of $C_2$. Write $1_{A_5}$ for the identity element of $A_5$. Case 1 provides you with the conjugating element $a$; then the homomorphism $f: C_2 \to A_5$ is defined by $f([0]) = 1_{A_5}$ and $f([1]) = a$.

I really don't see the point of non surjective bundles, just put the identity on some component if you really want to ignore them for some reason

@TedShifrin I think that's all for now :)

Not sure what that means in general, @Alessandro. You mean the trivial bundle, I suppose.

I mean the trivial rank 0 bundle

Well, my bundles cannot change rank from component to component.

Oh wait do you insist that bundles have constant rank?

LOL

Let's not have this discussion lol

Told you!

It's really pointless

Apparently, though, it started fiberless.

But we gave it Metamucil. (Old persons' joke.)

I don't get it, am I too young?

You are jumbling multiple things in that sentence. An automorphism $\varphi: N \to N$ is a bijective homomorphism. For $a \in N$, one may define an automorphism $\varphi_a: N \to N$ by $\varphi_a(n) = ana^{-1}$. An inner automorphism $\varphi$ is one of the form $\varphi_a$ for some $a \in N$.

Fiber supplements for (usually) older people.

lmao

Ah, makes sense

Apologies for disrupting Mike and Trisk's mathematics.

I don't know what $g$ is there. Way earlier $g$ was an element of $G \setminus N$, and here it's an element of $H$.

I always think that I understand semidirect products until I have to work with one. Then I need to spend 20 minutes to remember how they work, just to forget shortly after and having to that again the next time

$G \setminus N$ is not $H$, given for instance that it doesn't contain the identity element.

Reminds me of a question on an exam back in 1971. Show that a group cannot be a set-theoretic union of two subgroups.

At least, that's what I remember it to be.

Yeah, that's not true in any example, ever. You should write down an example.

@TedShifrin That's true

It seems to be what Mike just told Trisk to verify.

What do you think the expression $G \setminus N$ means?

There are a lot of weird results concerning groups written as a union of subgroups

As I recall, I did horribly on that particular exam ... shameful.

OK, we agree on what that means then. Write down an example (even a direct product) in which $H$ and $N$ are nontrivial groups, and verify that $G \setminus N \neq H \setminus \{0\}$, so your formula is wrong; in fact it is wrong for any extension $G$ of any groups $H$ and $N$ whatsoever so long as $H, N$ are nontrivial.

I think it would be most helpful for you if you wrote out all of these concepts very carefully from the start: automorphism of a group $G$; inner automorphism of a group $G$; semidirect products (which come from homomorphisms $\varphi: H \to \text{Aut}(N)$); then what it means for one of those homomorphisms $\varphi: H \to \text{Aut}(N)$ to be inner (that's what I was talking about above), and then why the corresponding semidirect products are trivial.

@Ted I found the paper I was thinking about

Oh cool. Not geometric enough for me, but interesting :P

There is also a not very well known lemma by Neumann saying that if $G$ is the union of finitely many (left) cosets of subgroups, then one of those subgroups has finite index

@TedShifrin I find it very weird how it stops at 7

Ah.

@Triskelion: Just to emphasize though ... Make sure you understand the difference between $G/N$ and $G\setminus N$. This seems to be hanging you up.

@TedShifrin Group theory: a geometric appro... Wait Gromov already did that

LOL ... thanks.

I did, too, in my book you all keep picking on. :P Well, a little bit.

Evening all

Geometric group theory is great (I'm writing my masters thesis in ggt and this is my completely unbiased opinion)

Hi @Edward

Evening, sir @Edward.

Grüß Gott!

Alessandro, you're always completely unbiased. We know that.

@TedShifrin I think that was perfectly appropriate

I would have used more colourful language

Well, I flagged and I presume some moderator did something appropriate.

Sometimes I wish I had such powers.

I will say that C-19 has precipitated an inordinate number of exam/homework questions posted on MSE. Maybe distance learning isn't ideal?

well, it's easier to cheat in most online scenarios

and a lot of teachers aren't properly aware either

Very naive teachers, yup. But some teachers haven't exactly been available for office hours and help, I bet. It's been about a month now, but I was helping someone with masters' level research because his adviser was accessible only by snail mail. Unbelievable.

why is a closed manifold the one without boundary?

what do we get? That the open disk, is a closed manifold?

No, "closed manifold" is what topologists say for compact manifold without boundary.

Just like we refer to a closed surface in multivariable calc ... That's the only time I actually do this.

yes, so the open disk is $D^2 - S^1$, is an open set, but a closed manifold?

is the open disk compact?

got me

It's always refreshing to know people read what I type in here.

right, compact is closed and bounded anyway..

ah, got it

a surface that closes in on itself

say a sphere

if it doesn't close in on itself, it's got boundary

Aha.

I hate sweating

any theo1ren to dry sweat quickly

@TedShifrin Hi Ted. Sorry if I bother you with my questions. But here it goes

In this section, you have the following functional

I[f], which is an integral

and then in the formula below we take the inf of I[h]

My question is WHY can we take the inf of a functional

Your loss function is likely non-negative by definition

p(x,y) is also nonnegative, so all your I[h] \ge 0

any set bounded below has an infimum

maybe you are asking why the infimum can be attained?

@nbro, why not? Observe that $I: f \mapsto \mathbb{R}$

so $\inf_{h \in H} I[h]$ can be read in plain English as the hypothesis, $h$, in the hypothesis set, $H$, that gives the least error, $I[h]$

define $f$ to be that hypothesis.

@Pig No, I am not asking that. This notation isn't clear to me

I was familiar with the inf of a set

but I[f] is not a set, it's an integral

ah thats what bothers you? The expression on wikipedia is shorthand for $\inf_{h \in H}\{I[h]\}$

remember, the integral evaluates to a real number, yes? meaning $\{I[h] | h \in H\}$ is a set of reals here

like Joe Shimo said, the set here is $\{I[h] | h \in H\}$, and they are taking infimum of this set

@JoeShmo But you can take the infimum of a set, but I[h] is not a set, but an integral

read my comment again

they are, in fact, taking the infimum over a set, not a single real number

It's always refreshing to know people read what I type in here.

- @TedShifrin

hahaha

@Pig exactly, that's what I was asking

you cannot take the infimum of I[f], because it's not a set

@JoeShmo {I[h]} makes 0 sense to me

I didn't say I[f]

I said $\{I[h]: h \in H\}$

but I understood that this is another shortcute for {I[h], such that h in H}

$\{I[h]\} = \{I[h] | h \in H\}$

Yes, I know that now

but the notation {I[h]} still makes absolutely zero sense to me

I wrote it out on the second comment up there

As I said many times in the past, the problem with mathematics is notation. Apart from that, math isn't complicated

lol

Ok, so let's summarise

we are taking the infimum with respect to h of a set of real numbers that represents the expected risk given h

right?

correct

you justify that that's the correct course of action by throwing in the Empirical Risk Minimization (ERM) "axiom"

ERM is a framework, really

But typically in machine learning we are looking for the hypothesis h and not the loss that we get if we use h

and that's all of machine learning on one leg

correct^

so $f = \inf_{h \in H} I[h]$ is a hypothesis

@Balarka @Alessandro opinions on Katatonia?

@JoeShmo You said it was a number

$\inf_{x \in X} (\cdot)$ gives back an element in $X$

Ha, ok

$I[h] \in \mathbb{R}$

But

not really, strictly speaking there's a typo there

it's either $f$ that satisfies $I[f] = inf_{h \in H} I[h]$

We are taking the infimum of {I[h] | h in H}, but all I[h] are numbers

or $f = arginf_{h \in H} I[h]$

so $I[h] \in \mathbb{R}$ gives us a set of real numbers to optimize over

Right

but that notation I[f] = inf I[h] is not standard to mean that inf will get a function

right?

but the optimization function ($\inf$) returns the hypothesis itself

yes it's not standard

stick to my notation will you

that's why people generally do $argmin$ (assuming min exists)

yes, strictly speaking, it's $f$, such that $I[f] = \inf I[h]$

technically what I said up there was wrong, $\inf I[h]$ returns $I[f]$, NOT $f$.

but that's how the Wikipedia article uses it..

If H is a topological subgroup with non empty interior, how do I show that its open?

2. As in 1, once you know that $G \to Aut(A_5) = S_5$ is onto, then because both left/right hand side has same order, it must be injective too, hence it's an isomorphism

Yes

If $H$ is normal subgroup $G$

@JoeShmo But why can inf also be used to denote arginf? Is this commonly used in math?

then for each $g \in G$, the map $c_g: H \to H$ defined by $h \to ghg^{-1}$ is always an automorphism of $H$

so you have a map $G \to Aut(H)$ which can be checked to be a homomorphism

by $g \to c_g$

It's both

That's too much effort for me :) Could be a good exercise for you if you need to write it up anyway

I mean you probably need to consider the union of those two sequences

It's a good question whether considering either of them is sufficient

i don't know off the top of my head, but i won't be surprised if that's the case

I vaguely remember some form of Butterfly lemma allows you to do some switching (or kernel and cokernel), but I don't know for sure. I'll leave it to group theory experts out there

Well they certainly can coincide

e.g. trivially, the direct product of them would admit a sequence of both types

well i'm not sure how $\phi$'s can be equal anyway, given that they have different domain/image

you probably only care about the group up to isomorphism

not the accompanying data (e.g. the extension itself)

@nbro it just so happens that in this case, $\inf = \arg\min$. This is because you're optimizing over a compact subset of the reals (closed and bounded). This is a topological detail for the purposes of machine learning. If you're interested, the principal at play here is the extreme value theorem -- en.wikipedia.org/wiki/Extreme_value_theorem

hmmm it's actually not super clear to me why $\min = \inf$ here to be honest. I would be surprised if this holds in the utmost generality (i.e. arbitrary loss, arbitrary probability distribution)

The claim using extreme value theorem is incorrect, because you are minimizing in the space of $h \in H$, not the space of $I[h]$

Yeah... I also don't know why you mentioned the extreme value theorem

The point is that you achieve a minimum in the set over which you are optimizing; in contrast to, say, an open interval $(0, 1)$, for which a "minimum" is $0$ (the greatest element $x \in \mathbb{R}$ for which $x \le t;\ \forall t \in (0,1)$), but $0 \not \in (0, 1)$.

I didn't even remember this theorem, to be honest

Hm actually I take back what I said, the problem is more that it's unclear $\{I[h]: h \in H\}$ is closed (or at least just attains the infimum)

yes, but, @Pig, $H$ is compact in practical application, and of course it doesn't work in utmost generality

sure - it's interesting when that would work though

it almost certainly works for convex loss, which is why people care about them

yup

but it's not clear at what generality it works

they don't really need more generality than that

its almost always convex optimization IIRC

i agree in practice lol