Representable homology classes on smooth manifolds

Question

Let $X$ be a closed (compact without boundary) smooth manifold. We can consider its singular homology $H_*(X,\mathbb{Z})$. Let $H_{k}(X,\mathbb{Z})$ be the $k$-th singular homology group of $X$ and let us consider $[\alpha]\in H_{k}(X,\mathbb{Z})$. I have seen in various places that it is sometimes possible to "represent" $[\alpha]$ as a smooth $k$-dimensional submanifold of $X$.

• How does this correspondence between elements in $H_{k}(X,\mathbb{Z})$ and $k$-dimensional submanifolds of $X$ exactly works?

• When exactly elements in $H_{k}(X,\mathbb{Z})$ can be represented by $k$-dimensional submanifolds?

• How is this related to intersection theory, the cup product in cohomology, Poincare duality and the intersection form of $X$ (when $X$ is even-dimensional)?

• Does this correspondence work for other homology theories?

Thanks.

Answer 1

Fundamental Classes

Let $Y$ be a connected, closed, orientable $k$-dimensional manifold, then $H_k(Y; \mathbb{Z}) \cong \mathbb{Z}$. The choice of a generator of $H_k(Y; \mathbb{Z})$ is equivalent to a choice of orientation. One often denotes the generator by $[Y]$; we call this a fundamental class of $Y$. If we consider $Y$ with it's other orientation, the corresponding fundamental class is $-[Y]$.

If we drop the orientability hypothesis, $H_k(Y; \mathbb{Z}) = 0$ so the above construction fails. However, $H_k(Y; \mathbb{Z}_2) \cong \mathbb{Z}_2$, so we define the $\mathbb{Z}_2$-fundamental class to be the generator of $H_k(Y; \mathbb{Z}_2)$.

We can also define the fundamental class if $Y$ is only compact instead of closed (i.e. $Y$ has boundary). In that case, we use the fact that $H_k(Y, \partial Y; \mathbb{Z}) \cong \mathbb{Z}$ in the orientable case and $H_k(Y, \partial Y; \mathbb{Z}_2) \cong \mathbb{Z}_2$ in the non-orientable case to define fundamental classes as relative homology classes.

All of the above is discussed in detail on the Manifold Atlas Project's fundamental class page.

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Submanifolds and Homology

Now suppose $Y$ is a connected, closed, orientable $k$-dimensional submanifold of $X$ and let $i : Y \to X$ denote the inclusion, then $i_* : H_k(Y; \mathbb{Z}) \to H_k(X; \mathbb{Z})$, in particular $i_*[Y] \in H_k(X; \mathbb{Z})$. Note, the element $i_*[Y]$ might be zero. For example, $S^1$ can be realised as a submanifold of $S^2$ via the equator, but $i_*[S^1] \in H_1(S^2; \mathbb{Z}) = 0$.

We can create homology classes on $X$ from different types of submanifolds as above using the different notions of fundamental class.

• If $Y$ is a non-orientable submanifold of $X$, using the $\mathbb{Z}_2$-fundamental class, $i_*[Y] \in H_k(X; \mathbb{Z}_2)$.
• If $X$ is a manifold with boundary and $Y$ is an orientable submanifold with boundary (so that $\partial Y \subseteq \partial X$), $i_*[Y, \partial Y] \in H_k(X, \partial X; \mathbb{Z})$.
• If $X$ is a manifold with boundary and $Y$ is a non-orientable submanifold with boundary, using the $\mathbb{Z}_2$-fundamental class, $i_*[Y, \partial Y] \in H_k(X, \partial X; \mathbb{Z}_2)$.

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Realisable Classes and The Steenrod Problem

An element $\alpha \in H_k(X; \mathbb{Z})$ is said to be realisable if there is a $k$-dimensional connected, closed, orientable $k$-dimensional submanifold $Y$ such that $\alpha = i_*[Y]$. The Steenrod Problem was whether every class is realisable. Thom showed in $1954$ that this is not true in general. However, it is true if $0 \leq k \leq 6$. What is true however is that for any class $\alpha$, there is an integer $N$ such that $N\alpha$ is realisable.

We could ask the analogue of Steenrod's Problem for $\mathbb{Z}_2$ coefficients: is every class in $H_k(X; \mathbb{Z}_2)$ realisable? The answer is yes!

I don't know about the analogue of Steenrod's Problem for manifolds with boundary.

Answer 2

These classes are not always representable, but classes of $H_l(M,Z/2)$ are representable. Positive multiple of elements of $H_l(M,Z)$ can be represented by smooth manifolds, this is shown in the thesis of Rene Thom.