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For an $n$-dimensional orientable closed manifold $M$, the simplicial volume is the infimum of the $l^1$-norm of the elements $\sum a_i \sigma_i$ ($a_i \in \mathbb{R}$) which represent the fundamental class. My question is: since $\sigma_i$ is a continuous map from the $n$-dimensional simplex to $M$, what does $a_i \sigma_i$ mean? Is it the same kind of map? I don't know how to make sense of this expression.

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    $\begingroup$ I edited your question to make it more clear; tell me if I've changed the meaning. $\endgroup$ – Ryan Reich May 21 '13 at 18:22
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    $\begingroup$ The definition of the singular chain complex is far from research level. $\endgroup$ – Fernando Muro May 21 '13 at 20:47
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It's only a formal (i.e. symbolic) sum, understood as an element of the $\mathbb R$-module generated by the simplexes $\sigma_i$ (similar to vectors from a basis in linear algebra), where addition is by definition component-wise, commutative and distributive. It does not possess a clear geometrical meaning. In particular it does not represent a map into $M$. Try to look for references on singular homology.

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    $\begingroup$ Perhaps you meant "references on singular homology". I would recommend Hatcher's book, which explains the distinctions between simplicial homology and singular homology. It is available at math.cornell.edu/~hatcher/AT/ATpage.html $\endgroup$ – Lee Mosher May 21 '13 at 18:12
  • $\begingroup$ @Lee : yes, sorry about the lapsus... I edited my answer. $\endgroup$ – Loïc Teyssier May 21 '13 at 18:41
  • $\begingroup$ Just consider the case when the communicative ring with unit is the real line R.The module is a map from R$ \times $ set of singular simplices to set of singular simplices.If ${a_i}{\sigma _i}$ is not singular simplex,how can it be a R-module?Since the fundamental class$\alpha $ is the generator of${H_n}(M;R) \cong R$.But for R,the generator can be any element except 0.So $c\alpha $ is the fundamental class when c$ \ne 0$?,thus the simplicial volume is 0 when we let c approach 0?I know it's wrong,but I don't know where $\endgroup$ – jiangsaiyin May 22 '13 at 10:03
  • $\begingroup$ @Lee:Just consider the case when the communicative ring with unit is the real line R.The module is a map from R$ \times $set of singular simplices to set of singular simplices.If${a_i}{\sigma _i}$is not singular simplex,how can it be a R-module?Since the fundamental class$\alpha $is the generator of${H_n}(M;R) \cong R$.But for R,the generator can be any element except 0.So$c\alpha $ is the fundamental class when c$ \ne 0$?,thus the simplicial volume is 0 when we let c approach 0?I know it's wrong,but I don't know where $\endgroup$ – jiangsaiyin May 22 '13 at 10:09
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As Lee suggests, you should look at Allen Hatcher's book. In particular, chapter 2 and section 3.3.

In practice, you can often think of $\sum a_{i}\sigma_{i}$ as coming from the following procedure. Let $\phi:X\rightarrow M$ be a degree $d$ map of $n$-manifolds. Then any singular cocycle (with integer coefficients) $\alpha$ representing the fundamental class of $X$ will push forward to a singular cocycle $\phi_{\ast}\alpha$ representing $d$ times the fundamental class of $M$. Thus, $\frac{1}{d}\phi_{\ast}\alpha$ will be a cocycle (with rational coefficients) representing the fundamental class of $M$.

The prototypical examples are as follows:

If $M$ is a torus, then there are covering maps $\phi: M\rightarrow M$ of arbitrarily high degree, hence $M$ has no simplicial volume. (We can iterate the above procedure, starting with any cocycle $\alpha$ representing the fundamental class of $M$, getting representatives $\alpha,\frac{1}{d}\phi_{\ast}(\alpha),\frac{1}{d^{2}}\phi_{\ast}^{2}(\alpha),\ldots$ with volume going to $0$.)

On the other hand, if $M$ is a higher genus surface, then this strategy is obstructed by Euler characteristic. Namely, if $\phi:Y\rightarrow M$ is a degree $d$ covering map, then $\chi(Y)=d\chi(X)$, so the genus of $Y$ grows linearly with $d$. In fact, you can show that $X$ has nontrivial simplicial volume using hyperbolic geometry.

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  • $\begingroup$ @David Cohen:Just consider the case when the communicative ring with unit is the real line R.The module is a map from R $ \times $ set of singular simplices to set of singular simplices.If${\alpha _i}{\sigma _i}$is not singular simplex,how can it be a R-module?Since the fundamental class $\alpha $ is the generator of${H_n}\left( {M;R} \right) \cong R$.But for R,the generator can be any element except 0.So$c\alpha $is the fundamental class when c$ \ne 0$ ?,thus the simplicial volume is 0 when we let c approach 0?I know it's wrong,but I don't know where – jiangsaiyin 15 hours ago $\endgroup$ – jiangsaiyin May 23 '13 at 2:04
  • $\begingroup$ I'm not sure what you mean by module, but I see where you are getting confused. There is a map $H_{n}(M,\mathbb{Z})\rightarrow H_{n}(M,R)$ defined exactly the way you think it would be (the image of $\sum a_{i}\sigma_{i}$ is just $\sum a_{i}\sigma_{i}$, since integers are also real numbers.) The fundamental class in $H_{n}(M,\mathbb{Z})$ is the generator of this group, call it $[M]$. The fundamental class in $H_{n}(M,R)$ is just the image $[M]$ under the map I mentioned above. $\endgroup$ – David Cohen May 24 '13 at 17:40
  • $\begingroup$ It may seem that any cycle representing this class will always have integer coefficients, but this is not true! For instance, consider the circle $S^{1}$, thought of as the unit circle in the complex plane. Then the fundamental class is represented by the simplex $\sigma:[0,1]\rightarrow S^{1}$ given by $x\mapsto e^{2\pi i x}$. Clearly, $\sigma$ has integer coefficients (the only coefficient is $1$.) However, in $H_{1}(S^{1};\mathbb{R})$, this is equal (cohomologous) to $\frac{1}{2}\tau$ where $tau$ is the simplex $\tau:x\mapsto e^{4\pi i x}$. It is a good exercise to work out why. $\endgroup$ – David Cohen May 24 '13 at 17:48

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