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Stream: practice: communication

Topic: Professor doesn't accept that he is wrong

Jean-Baptiste Vienney (Jan 14 2025 at 15:39):

I'm taking a course on representation theory this semester.

The professor said that a representation of a group $G$ over a field $F$ is equivalent to a left module $V$ over the ring $F[G]$ such that $(\lambda s) \cdot x=s \cdot (\lambda x)$ for all $\lambda \in F$, $s \in F[G]$ and $x \in V$ (at first he said $(\lambda g) \cdot x=g \cdot (\lambda x)$ for all $\lambda \in F$, $g \in G$ and $x \in V$).

I sent an email to the professor telling him that this additional requirement is automatically verified. The prof doesn't believe me. I sent him a proof that the identity $(\lambda g) \cdot x=g \cdot (\lambda x)$ for all $\lambda \in F$, $g \in G$ and $x \in V$ for all $\lambda \in F$, $g \in G$ and $x \in V$ (the additional requirement he wrote the first time he talked about $F[G]$-modules) is automatically verified by every $F[G]$-module. I think he didn't read my proof. He said that he will send me his comments but never did it. He gave me a "counterexample". I explained to him why it is wrong but it seems that he doesn't trust me or doesn't understand what I say. He said to the class today that I discussed this thing with him but also says to the class that it's easy to come with counterexamples of $F[G]$-modules which don't verify the additional axiom.

I start wanting to insult the prof in front of all the class but I don't think I should do this. What should I do? I don't know if either the prof doesn't accept that he is wrong or doesn't understand the issue. Maybe he thinks that he is much more knowledgeable than me since he is a university professor doing research on the Langlands program blablabla and he is not able to trust a student telling him that he is wrong, even when the student gives a formal mathematical proof of what he states.

It makes me want to drop the course or go do mathematics alone in the mountains far from civilization and universities.

Patrick Nicodemus (Jan 14 2025 at 18:40):

I don't know the guy, but I think you might be being too hard on him. Before we talk about the personal dynamics, can you convince me this is true? I can't see why it's true.

This seems subtle to me because the category of $F$-modules is a subcategory of the category of $\mathbb{Z}$-modules, but the tensor product of $\mathbb{Z}$-modules won't necessarily agree with the tensor product of $F$-modules in general, it's not a strong monoidal inclusion.

Basically what I'm getting at is that the diagrams for a monoid and an algebra for the monoid could be drawn in two different categories here, and have two slightly different meanings in each case. It seems plausible to me that the professor is saying that "left module over an F-algebra" is stronger than "left module over the underlying ring of an F-algebra". In one setting the action has to be a map $A \otimes M \to M$, in the other we want a map $A\otimes_F M \to M$.

Jean-Baptiste Vienney (Jan 14 2025 at 18:43):

Ok, so I give you the proof of the statement.

Jean-Baptiste Vienney (Jan 14 2025 at 18:43):

Let $V$ be an $F[G]$-module where $F$ is a field and $G$ a group.

Jean-Baptiste Vienney (Jan 14 2025 at 18:44):

For every $\lambda \in F$ and $x \in V$, we can write $\lambda x:=(\lambda e)x$ where $e$ is the identity of $G$.

Jean-Baptiste Vienney (Jan 14 2025 at 18:45):

For every $\lambda \in F$ and $s=\underset{g \in G}{\sum}\mu_g g \in F[G]$, we can define $\lambda s \in F[G]$ by $\lambda s:=\underset{g \in G}{\sum}\lambda\mu_g g$.

Jean-Baptiste Vienney (Jan 14 2025 at 18:48):

Now I claim that for every $\lambda \in F$, $s \in F[G]$ and $x \in V$, we have:
$(\lambda s)x=s(\lambda x)$.

Patrick Nicodemus (Jan 14 2025 at 18:49):

Hold on one minute

Patrick Nicodemus (Jan 14 2025 at 18:50):

Can you identify or point to an axiom which gives us λx=(λe)x?

Jean-Baptiste Vienney (Jan 14 2025 at 18:50):

The point is that $\lambda x$ doesn't have any meaning at first, since $\lambda$ is not an element of $F[G]$.

Patrick Nicodemus (Jan 14 2025 at 18:50):

Oh, it's just notation, I see.

Jean-Baptiste Vienney (Jan 14 2025 at 18:51):

Yeah, and I think this is where the prof starts to be confused.

Jean-Baptiste Vienney (Jan 14 2025 at 18:52):

An $F[G]$-module is an $F$-vector space if you identify $F$ with $Fe$.

Patrick Nicodemus (Jan 14 2025 at 18:52):

Okay. You can continue your proof.

Jean-Baptiste Vienney (Jan 14 2025 at 18:53):

So, write $s=\underset{g \in G}{\sum}\mu_g g$.

John Baez (Jan 14 2025 at 18:53):

@Jean-Baptiste Vienney - I'd say: don't do anything, unless the professor is open to a reasonable conversation like you're having with Patrick.

There are times when people are unreasonable and there is nothing good to do about it: any further struggle will only make the situation worse. I try to take these times as opportunities to improve my self-control and sense of calm detachment. I got many such opportunities when I moderated a newsgroup on physics where crackpots would present their theories. In some ways it's even more infuriating when the offender is a professor, but I think the principles are the same: when someone's ego is at stake, they may not admit being wrong, and they'll interpret any further attempt to convince them as an attack that needs to be fought.

The worst possible thing to do is insult them, because this convinces them that they're correct in believing that you are attacking them. Often the best thing to do is nothing. You can triumph by not getting pulled into an emotional whirlpool about what is ultimately a rather small issue in the grand scheme of life.

Jean-Baptiste Vienney (Jan 14 2025 at 18:55):

Thank you, resignating myself should indeed be the best thing to do. It's easier now that someone agrees with me.

Jean-Baptiste Vienney (Jan 14 2025 at 18:56):

(because now I'm sure that I'm not crazy...)

Jean-Baptiste Vienney (Jan 14 2025 at 18:57):

I'll continue the proof for Patrick

Jean-Baptiste Vienney (Jan 14 2025 at 19:00):

We have:
$s(\lambda x)=(\underset{g \in G}{\sum}\mu_g g)(\lambda x)=(\underset{g \in G}{\sum}\mu_g g)((\lambda e)x)=((\underset{g \in G}{\sum}\mu_g g)(\lambda e))x=(\underset{g \in G}{\sum}\mu_g\lambda g)x=(\underset{g \in G}{\sum}\lambda\mu_g g)x=((\lambda e)(\underset{g \in G}{\sum}\mu_g g))x=\lambda(sx)$.

Patrick Nicodemus (Jan 14 2025 at 19:03):

Okay, I think I still need to clear something up.
Does your professor assume explicitly that $V$ is already an $F$-vector space?

Patrick Nicodemus (Jan 14 2025 at 19:04):

Like, in the right-to-left. Is it:
"Let $V$ be an $F$-vector space. If $V$ is a left module of $F[G]$, and (...), then $V$ has a $G$-representation"
or
"Let $V$ be an arbitrary set. If $V$ is a left module of $F[G]$, and (...), then $V$ has a $G$-representation"

Jean-Baptiste Vienney (Jan 14 2025 at 19:09):

No, he doesn’t assume that $V$ is a vector space…

Jean-Baptiste Vienney (Jan 14 2025 at 19:10):

This is the second option.

Jean-Baptiste Vienney (Jan 14 2025 at 19:11):

Let me check what he wrote exactly

Jean-Baptiste Vienney (Jan 14 2025 at 19:12):

From my notes of the course:

Jean-Baptiste Vienney (Jan 14 2025 at 19:12):

Screenshot 2025-01-14 at 2.12.21 PM.png

Patrick Nicodemus (Jan 14 2025 at 19:22):

Okay. Well, it doesn't explicitly say that $V$ is a vector space. But we could interpret the fact that he wrote $\lambda v$ as implying that, right? If I read $\lambda v$, I automatically insert the phrase "Let $V$ be a vector space" at the beginning of 3.5. Do you agree with that?

Jean-Baptiste Vienney (Jan 14 2025 at 19:29):

When I talked with the prof he said that an $FG$-module $V$ is a vector space by identifying $F$ with $Fe \subseteq FG$.

Patrick Nicodemus (Jan 14 2025 at 19:44):

Okay. I think my position is that if you interpret $V$ as being, a priori, an $F$ vector space, then you should explicitly assume that the the action of $FG$ on $V$ is compatible with the existing action. If $V$ is not a vector space a priori and the vector space structure is derived from the action of $FG$, then it's automatically compatible and you're right.

Jean-Baptiste Vienney (Jan 14 2025 at 21:56):

I've just realized that the prof answered my last message but I just saw the first line of it on my phone when he sent it on Friday. So I'm answering this message where he still doesn't agree with me...

Jean-Baptiste Vienney (Jan 14 2025 at 21:57):

Hopefully, we'll finally solve the issue.

Jean-Baptiste Vienney (Jan 14 2025 at 21:57):

I asked him if he considers $V$ as a vector space as suggested by Patrick.

Morgan Rogers (he/him) (Jan 15 2025 at 16:50):

It might be faster for us to see his purported counterexample?

Jean-Baptiste Vienney (Jan 15 2025 at 18:24):

Sure, he says that if $V$ is a complex vector space, then it is a $F[\{e\}]-module$ with $(\lambda e).x:=\overline{\lambda}x$. ($F:=\mathbb{C}$)

Jean-Baptiste Vienney (Jan 15 2025 at 18:24):

And then his additional requirement is not satisfied

Jean-Baptiste Vienney (Jan 15 2025 at 18:26):

But it makes sense only if you consider $V$ as vector space and an $FG$-module independently and interpret $\lambda x$ as being the result of the action of $F$ on $V$ as a vector space.

Jean-Baptiste Vienney (Jan 15 2025 at 18:26):

Which is not what he did in the lectures.

Jean-Baptiste Vienney (Jan 15 2025 at 18:28):

My conjecture now is that he was not very sure whether he considers $V$ as a vector space + an $FG$-module or just as an $FG$-module in his definition.

Jean-Baptiste Vienney (Jan 16 2025 at 15:03):

The prof told me today that in the course $V$ is always an $F$-vector space so now we agree on everything :)

Patrick Nicodemus (Jan 16 2025 at 15:27):

I'm glad you resolved it.
Until we get to the point where all mathematics is formalized in Coq as a matter of course, there will be ambiguity in communication in mathematics, and we will have to deal with these frustrating communications breakdowns. Just something to deal with

Paolo Perrone (Jan 16 2025 at 15:32):

I guess it's also one more instance of the fact that it matters in what category we are working.

Morgan Rogers (he/him) (Jan 16 2025 at 17:13):

Patrick Nicodemus said:

Until we get to the point where all mathematics is formalized in Coq as a matter of course, there will be ambiguity in communication in mathematics, and we will have to deal with these frustrating communications breakdowns.

1. Were you joking, or do you actually hope for maths to be formalized in Coq (including those appearing in lectures, as is the case in this discussion)?
2. What the prof originally said was not that ambiguous, in the end. @Jean-Baptiste Vienney was right, in that V automatically gets a vector space structure independently of assuming it has one (and it seems bizarre to assume that it would come with an incompatible vector space structure?) but a sensible inference of the missing data that makes the definition type-check corrects it.

John Baez (Jan 16 2025 at 17:45):

I hope nobody seriously thinks formalizing math is a substitute for talking to other people until we understand each other and agree on mathematical facts.

Patrick Nicodemus (Jan 22 2025 at 18:07):

Yes, I was joking.

John Baez (Jan 22 2025 at 18:08):

Whew! Some people are so crazy about formalization that I couldn't tell (since I don't know you well enough).