Steenrod problem

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1 Introduction

Given a space X, there is a homomorphism $\Phi : \Omega^{SO}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z})$${{Stub}} == Introduction == <wikitex>; Given a space X, there is a homomorphism $\Phi : \Omega^{SO}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z})$, called the Thom homomorphism, given by $[M,f] \to f_{*}([M])$ where $[M]$ is the fundamental class of $M$. The elements in the image of $\Phi$ are called representable. In certain situations it is convenient to assume that a homology class is representable. In dimensions <script type="math/tex">\Phi : \Omega^{SO}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z})$, called the Thom homomorphism, given by $[M,f] \to f_{*}([M])$$[M,f] \to f_{*}([M])$ where $[M]$$[M]$ is the fundamental class of

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$M$. The elements in the image of $\Phi$$\Phi$ are called representable.

In certain situations it is convenient to assume that a homology class is representable. In dimensions $0$$0$ and $1$$1$ it is clear that $\Phi$$\Phi$ is surjective (even an isomorphism). It is less obvious in dimension $2$$2$, but also can be shown geometrically. This made Steenrod raise his famous problem in 1946 [Eilenberg1949]:

Given a simplicial complex $X$$X$, is every (integral) homology class representable?

The answer was given by Thom in 1954 [Thom1954]. He showed that in dimensions $0 \leq m \leq 6$$0 \leq m \leq 6$ this is true but in general this is not the case. He constructed a counter example in dimension $m=7$$m=7$. Thom also showed that the corresponding problem with $\mathbb{Z}/2$$\mathbb{Z}/2$ coefficients is correct, that is the corresponding homomorphism $\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$$\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$ is surjective. Thom also proved the following:

Theorem 1.1 [Thom1954, Theorem III.4]. For every class in dimension $p$$p$ of integral homology of a finite polyhedron K, there exists a non zero integer $N$$N$ which depends only on p, such that the product $Nz$$Nz$ is the image of the fundamental class of a closed oriented differentiable manifold.

More about that can be found in [Sullivan2004].

2 References

• [Eilenberg1949] S. Eilenberg, On the problems of topology, Ann. of Math. (2) 50 (1949), 247–260. MR0030189 (10,726b) Zbl 0034.25304
• [Sullivan2004] D. Sullivan, René Thom's work on geometric homology and bordism, Bull. Amer. Math. Soc. (N.S.) 41 (2004), no.3, 341–350 (electronic). MR2058291 Zbl 1045.57001
• [Thom1954] R. Thom, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28 (1954), 17–86. MR0061823 (15,890a) Zbl 0057.15502

$ and \Phi : \Omega^{SO}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}), called the Thom homomorphism, given by $[M,f] \to f_{*}([M])$$[M,f] \to f_{*}([M])$ where $[M]$$[M]$ is the fundamental class of

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$M$. The elements in the image of $\Phi$$\Phi$ are called representable.

In certain situations it is convenient to assume that a homology class is representable. In dimensions $0$$0$ and $1$$1$ it is clear that $\Phi$$\Phi$ is surjective (even an isomorphism). It is less obvious in dimension $2$$2$, but also can be shown geometrically. This made Steenrod raise his famous problem in 1946 [Eilenberg1949]:

Given a simplicial complex $X$$X$, is every (integral) homology class representable?

The answer was given by Thom in 1954 [Thom1954]. He showed that in dimensions $0 \leq m \leq 6$$0 \leq m \leq 6$ this is true but in general this is not the case. He constructed a counter example in dimension $m=7$$m=7$. Thom also showed that the corresponding problem with $\mathbb{Z}/2$$\mathbb{Z}/2$ coefficients is correct, that is the corresponding homomorphism $\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$$\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$ is surjective. Thom also proved the following:

Theorem 1.1 [Thom1954, Theorem III.4]. For every class in dimension $p$$p$ of integral homology of a finite polyhedron K, there exists a non zero integer $N$$N$ which depends only on p, such that the product $Nz$$Nz$ is the image of the fundamental class of a closed oriented differentiable manifold.

More about that can be found in [Sullivan2004].

2 References

• [Eilenberg1949] S. Eilenberg, On the problems of topology, Ann. of Math. (2) 50 (1949), 247–260. MR0030189 (10,726b) Zbl 0034.25304
• [Sullivan2004] D. Sullivan, René Thom's work on geometric homology and bordism, Bull. Amer. Math. Soc. (N.S.) 41 (2004), no.3, 341–350 (electronic). MR2058291 Zbl 1045.57001
• [Thom1954] R. Thom, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28 (1954), 17–86. MR0061823 (15,890a) Zbl 0057.15502

$ it is clear that $\Phi$ is surjective (even an isomorphism). It is less obvious in dimension $, but also can be shown geometrically. This made Steenrod raise his famous problem in 1946 {{cite|Eilenberg1949}}: Given a simplicial complex $X$, is every (integral) homology class representable? The answer was given by Thom in 1954 {{cite|Thom1954}}. He showed that in dimensions $\Phi : \Omega^{SO}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z})$, called the Thom homomorphism, given by $[M,f] \to f_{*}([M])$$[M,f] \to f_{*}([M])$ where $[M]$$[M]$ is the fundamental class of

Tex syntax error

$M$. The elements in the image of $\Phi$$\Phi$ are called representable.

In certain situations it is convenient to assume that a homology class is representable. In dimensions $0$$0$ and $1$$1$ it is clear that $\Phi$$\Phi$ is surjective (even an isomorphism). It is less obvious in dimension $2$$2$, but also can be shown geometrically. This made Steenrod raise his famous problem in 1946 [Eilenberg1949]:

Given a simplicial complex $X$$X$, is every (integral) homology class representable?

The answer was given by Thom in 1954 [Thom1954]. He showed that in dimensions $0 \leq m \leq 6$$0 \leq m \leq 6$ this is true but in general this is not the case. He constructed a counter example in dimension $m=7$$m=7$. Thom also showed that the corresponding problem with $\mathbb{Z}/2$$\mathbb{Z}/2$ coefficients is correct, that is the corresponding homomorphism $\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$$\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$ is surjective. Thom also proved the following:

Theorem 1.1 [Thom1954, Theorem III.4]. For every class in dimension $p$$p$ of integral homology of a finite polyhedron K, there exists a non zero integer $N$$N$ which depends only on p, such that the product $Nz$$Nz$ is the image of the fundamental class of a closed oriented differentiable manifold.

More about that can be found in [Sullivan2004].

2 References

• [Eilenberg1949] S. Eilenberg, On the problems of topology, Ann. of Math. (2) 50 (1949), 247–260. MR0030189 (10,726b) Zbl 0034.25304
• [Sullivan2004] D. Sullivan, René Thom's work on geometric homology and bordism, Bull. Amer. Math. Soc. (N.S.) 41 (2004), no.3, 341–350 (electronic). MR2058291 Zbl 1045.57001
• [Thom1954] R. Thom, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28 (1954), 17–86. MR0061823 (15,890a) Zbl 0057.15502

\leq m \leq 6$ this is true but in general this is not the case. He constructed a counter example in dimension $m=7$. Thom also showed that the corresponding problem with $\mathbb{Z}/2$ coefficients is correct, that is the corresponding homomorphism $\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$ is surjective. Thom also proved the following: {{beginthm|Theorem|{{cite|Thom1954|Theorem III.4}}}} For every class in dimension $p$ of integral homology of a finite polyhedron K, there exists a non zero integer $N$ which depends only on p, such that the product $Nz$ is the image of the fundamental class of a closed oriented differentiable manifold. {{endthm}} More about that can be found in {{cite|Sullivan2004}}. == References == {{#RefList:}} [[Category:Theory]]\Phi : \Omega^{SO}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}), called the Thom homomorphism, given by $[M,f] \to f_{*}([M])$$[M,f] \to f_{*}([M])$ where $[M]$$[M]$ is the fundamental class of

Tex syntax error

$M$. The elements in the image of $\Phi$$\Phi$ are called representable.

In certain situations it is convenient to assume that a homology class is representable. In dimensions $0$$0$ and $1$$1$ it is clear that $\Phi$$\Phi$ is surjective (even an isomorphism). It is less obvious in dimension $2$$2$, but also can be shown geometrically. This made Steenrod raise his famous problem in 1946 [Eilenberg1949]:

Given a simplicial complex $X$$X$, is every (integral) homology class representable?

The answer was given by Thom in 1954 [Thom1954]. He showed that in dimensions $0 \leq m \leq 6$$0 \leq m \leq 6$ this is true but in general this is not the case. He constructed a counter example in dimension $m=7$$m=7$. Thom also showed that the corresponding problem with $\mathbb{Z}/2$$\mathbb{Z}/2$ coefficients is correct, that is the corresponding homomorphism $\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$$\Phi : \Omega^{O}_{\ast}(X) \to H_{\ast}(X,\mathbb{Z}/2)$ is surjective. Thom also proved the following:

Theorem 1.1 [Thom1954, Theorem III.4]. For every class in dimension $p$$p$ of integral homology of a finite polyhedron K, there exists a non zero integer $N$$N$ which depends only on p, such that the product $Nz$$Nz$ is the image of the fundamental class of a closed oriented differentiable manifold.

More about that can be found in [Sullivan2004].

2 References

• [Eilenberg1949] S. Eilenberg, On the problems of topology, Ann. of Math. (2) 50 (1949), 247–260. MR0030189 (10,726b) Zbl 0034.25304
• [Sullivan2004] D. Sullivan, René Thom's work on geometric homology and bordism, Bull. Amer. Math. Soc. (N.S.) 41 (2004), no.3, 341–350 (electronic). MR2058291 Zbl 1045.57001
• [Thom1954] R. Thom, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28 (1954), 17–86. MR0061823 (15,890a) Zbl 0057.15502