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This question can be too broad if we're not careful. It's my hope that an answer would consist of a short list of theorems and their dependencies. If this is too much to ask or is in some way inappropriate, then I'm sorry.

In Springer's book, "Linear Algebraic Groups (Second Edition)", we have the following theorem:

Theorem 6.2.7(iii): Two Borel subgroups of [a linear algebraic group] are conjugate.

NB: We're working over an algebraically closed field.

The proof given is:

If $B$ and $B'$ are two Borel subgroups then $B'$ is conjugate to a subgroup of $B$ and $B$ to a subgroup of $B'$. Hence $\dim B=\dim B'$, which implies (iii).

Context:

I'm learning the basics of linear algebraic group theory, to facilitate research in the area. My supervisor suggested working backwards from a couple of theorems (to lighten the load), one of which being Theorem 6.2.7(iii).

Currently, I am halfway through Chapter 3 of Springer's book. I have copies of Humphreys' and Borel's books, which I dive into to strengthen my understanding of what's covered in Springer's.

My supervisor and I discuss the general theory once a week. We're building up my knowledge & intuition slowly but surely.

The Question:

What are the main theorems Theorem 6.2.7(iii) relies upon, and what do they rely upon, and so on until the basics are covered? Where can these theorems be found in Springer's book?

Thoughts:

The proof given above can be broken down into three statements I don't yet understand:

  1. If $B$ and $B'$ are two Borel subgroups, then $B$ is conjugate to a subgroup of $B'$.

  2. If $B$ and $B'$ are conjugate to subgroups of each other, then $\dim B=\dim B'$.

  3. If $\dim B=\dim B'$, then $B$ and $B'$ are conjugate to each other.

The one that makes the most sense to me is 3. However, I don't know where any of them is in Springer's book.

Further Context:

To get some idea of my ability with linear algebraic groups, see the following questions of mine:

This feels like trying to run before I can walk; however, I am sure many professional mathematicians, when reading lengthy papers/books, work backwards from theorems they deem useful, therefore avoiding having to read entire documents. I hope to develop this skill. I just need a leg-up for now, please.

I have included the tag because it approximates what I am after nicely.

Shaun
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    Are you working over an algebraically closed field? – Moishe Kohan Apr 25 '23 at 14:42
  • Sorry, yes; I'll add that detail now. – Shaun Apr 25 '23 at 14:48
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    You should read the whole proof, instead of just the part about (iii). In the beginning Springer shows that if B is a Borel, then every parabolic subgroup contains a conjugate of B. Then he shows that Borel subgroups are parabolic. Hence the result. – spin Apr 28 '23 at 08:42
  • That's progress, @spin. Thank you. It's not entirely an answer to my question though. What theorems does that part of the proof rely on? I know they use the Borel's fixed point theorem, but what does that rely on, etc.? – Shaun Apr 28 '23 at 12:28
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    @Shaun: Why not just read the book and work backwards? This is a normal and common way to read math when you want to understand a single result without reading the whole text from the beginning to the end. I think Springer states pretty clearly in each proof what statements he is using. So for example you can assume Borel fixed pt thm holds, and postpone the proof of that until you understand how to use it to get to your goal (conjugacy of Borels). – spin Apr 28 '23 at 15:02
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    Read the proofs and see try to see what results are used, and what concepts come up. Eg. to understand the Borel fixed point thm, you need to know what a complete variety is, and you need to understand how parabolic subgroups are characterized in terms of the quotient variety $G/P$. If you get stuck look at other standard books or papers where the same result is proven, for example Humphreys' book on linear algebraic groups. – spin Apr 28 '23 at 15:06

1 Answers1

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Here is how you prove 2 and 3 (assuming only that the subgroups are Zariski-closed):

  1. Iterating the conjugations $f: B\to B', f': B'\to B$ you obtain a sequence of closed subgroups $$ B=B_1\supset B_2\supset B_3.... $$ all conjugate to each other. But such a sequence has to stabilize. From this you see that $f(B)=B'$.

  2. Even though it is an overkill, there is a more general fact that you should know (the Ax–Grothendieck theorem): If $X$ is a variety over an algebraically closed field then every injective morphism $X\to X$ is also surjective.

Moishe Kohan
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  • Thank you (+1). This definitely helps, but it's not what I asked for. I'm after a list of theorems from Springer's book. – Shaun Apr 25 '23 at 15:38
  • @Shaun: Why? This is a very odd request. Do you want to learn the material or do you want to learn a list of theorem numbers? – Moishe Kohan Apr 25 '23 at 15:42
  • I am a bit confused about the first part. Why is that even necessary? If $B$ is conjugate to a subgroup of $B'$, say via the map $f:B\rightarrow B'$ by your notation, then i) $\dim f(B)\leq \dim B'$ and, since conjugation is an isomorphism, ii) $\dim B=\dim f(B)$. Now we do the same for $B', f'$ and we get the equality, no? – Levent Apr 25 '23 at 15:43
  • That's a fair point. I ask because I want to look up their proofs and really understand them thoroughly, from the ground up. – Shaun Apr 25 '23 at 15:44
  • @Levent: I am proving a stronger result, but, yes, just for the equality of dimensions one iteration suffices. – Moishe Kohan Apr 25 '23 at 15:52
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    Oh, I see, you are proving 2 and.3 together. – Levent Apr 25 '23 at 15:54
  • @Shaun: My suggestion is still to focus on the content. As you go on, you can make your own list of useful theorems (with numbers) from the book. – Moishe Kohan Apr 25 '23 at 15:59