Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a real connected orientable closed $n$-dimensional compact differentiable manifold.

A connected oriented closed $d$-dimensional submanifold $i:M\to X$ (i.e. $M$ is a real connected orientable closed compact differentiable manifold and $i$ is a topological embedding) has a fudamental class $[M]\in H_d(M,\mathbb{Z})$. This can be considered as an element $i_*([M])$ of the singular homology $H_d(X,\mathbb{Z})$.

  1. The accepted answer to this question states, that a multiple $\lambda x$ of every element $x\in H_i(X,\mathbb{Z})$ for $0\leq i\leq n$ is of the form $i_*([M])$ for some $M$. If $n\leq 8$ then one may chose $\lambda=1$.
  2. An answer to this question however states that for example $2[M]$ is for $M=S^1$ not representable in this way. Also the other answers seem to be in conflict with (1.).

Where is my misunderstanding? Is the first answer not about embeddings $i:M\to X$ but about immersions $i$ or arbitrary continuous maps?

share|improve this question
2  
This is a very good question! –  Georges Elencwajg Jan 19 '13 at 13:25
add comment

1 Answer

up vote 4 down vote accepted

Thom's quoted article in the link to 1. certainly does not contain the "result" that you cite, which is false. The accepted answer you mention is at least ambiguously formulated.

Indeed, Don Stanley, who gave an answer to the MO question linked to your 2. is perfectly right: the homology class $n[M]\in H_n(M,\mathbb Z)$ is certainly not represented by a closed submanifold of $M$ as soon as $n\geq 2$.
The reason is that $H_n(M,\mathbb Z)$ is canonically isomorphic to $\mathbb Z$ once an orientation has been fixed and thus has no torsion: we cannot have $n[M]=[M]$, and obviously the only closed manifold of $M$ that can represent a homology class in $H_n(M,\mathbb Z)$ is $M$, which represents only $1\cdot [M]$.

share|improve this answer
1  
In Thom's theorem the case when $i=n$ is excluded from the theorem, but the rest of the cases remain true, if I understood it correctly. –  user17786 Jan 19 '13 at 13:45
    
Dear @user17786, I don't understand what you mean. Thom has proved many theorems in the article, so please give a precise reference to it, since everything Thom wrote there is true (as far as I know) without having to exclude any case. –  Georges Elencwajg Jan 19 '13 at 18:17
    
Sorry for the late answer. Magneto showed a contradiction among two answers. I was trying to explain why the first claim does not follow from Thom's paper (you have explained why the second answer is correct, but I think it would also be interesting to know what is the precise content of Thom's theorem and why the first claim does not follow) Theorem II.1 assumes that the homology classes are not in the top dimension, and theorem II.29 (which is the one on submanifolds representing integral homology classes) uses it, although it does not say explicitly that the top dimension is excluded. –  user17786 Jan 28 '13 at 21:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.