Symmetric functions possess a common zero.

Question

So possess a common zero, means $\exists x \in S^k: f_i(x) = 0$, right?

Then I could not follow this brief proof - what is the corollary? Because the information in the proof is so little, I couldn't even guess which.

Thanks =)

Theorem. Any $k$ smooth functions $f_1,\ldots, f_k$ on $S^k$ that satisfy the symmetry condition $f_i(-x)=-f_i(x)$, $i=1,\ldots, k$, must possess a common zero.

Proof. If not, apply the corallary to the map $$ f(x)=(f_1(x),\ldots, f_k(x),0),$$ taking the $x_{k+1}$ axis for $l$.

Edit: the corollary turns out to be the theorem mentioned in this question.

The proof mentions a corollary ... presumably just a fewl lines above the theorem statement. — Hagen von Eitzen, Jul 18 '13 at 06:07
The corollary is the result called "Theorem" right before it that I answered your question about earlier. — Henry T. Horton, Jul 18 '13 at 06:10
Hi @HagenvonEitzen, I didnt' know. But according to Henry's reply, the corollary should be this: http://i.stack.imgur.com/m7Rw7.png — 1LiterTears, Jul 18 '13 at 06:27
Hi @julien, I'm sorry.. But I didn't know... should be this one though: http://i.stack.imgur.com/m7Rw7.png — 1LiterTears, Jul 18 '13 at 06:29
@HagenvonEitzen Then that is exactly Borsuk-Ulam, right? Nope, I guess it is the one right before. — 1LiterTears, Jul 18 '13 at 06:33
Great, got it, thanks @Julien. Though you comment really is an answer.. They call it corollary, that's why confused me. GP alters names a lot, I had other questions aroused by so before.. =) — 1LiterTears, Jul 18 '13 at 06:41
Dear Jellifish, please edit the question to contain all relevant information, like what the corollary you are referring to it. Honestly, by now you should know better... I will probably start ignoring your questions :-/ — Mariano Suárez-Álvarez, Jul 18 '13 at 06:55
Dear @MarianoSuárez-Alvarez: First apologize that I didn't put the corollary as soon as I realized it. However, my question start with "what is the corollary, I can't find it." :_( — 1LiterTears, Jul 18 '13 at 07:17
And please please don't ignore my question, because the great @MarianoSuárez-Alvarez has been so knowledgeable and helpful :_(... — 1LiterTears, Jul 18 '13 at 07:21
Oh no.. What should I do.. I'll be more careful next time, good @MarianoSuárez-Alvarez... — 1LiterTears, Jul 18 '13 at 07:26

score 2 · Accepted Answer · answered Jul 18 '13 at 07:06

2

Yes, with this theorem (why do they call it corollary, then...?), it works exactly like they say, by contradiction. If the $f_j$'s don't vanish simultaneously, then $f$ takes $S^k$ to $\mathbb{R}^{k+1}\setminus\{0\}$, and satisfies the symmetry condition. So by the corollary-theorem, it must intersect in particular the line $\{x_1=\ldots=x_k=0\}$. But this gives a simultaneous zero. Contradiction.

answered Jul 18 '13 at 07:06

Julien

44,791

Thanks julien for your help with the answer and edit to the question! – 1LiterTears Jul 18 '13 at 07:22

Symmetric functions possess a common zero.

1 Answers1