# Representable homology classes on smooth manifolds

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.

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.

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, $i_*[Y, \partial Y] \in H_k(X, \partial X; \mathbb{Z}_2)$.

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.

• I have to run now but I will try to expand on my answer later. – Michael Albanese Apr 19 '16 at 23:11
• Thanks for the detailed answer. In his thesis Thom says that every $\mathbb{Z}_{2}$-class in dimension less than half of the dimension of the manifold is realisable. Was this result improved later? Also, do you know of any reference where all these issues are explained in detail? (in particular how to realize the cup product in cohomology in terms of interesection of the submanifolds of the corresponding homology classes, assuming they are realisable). – Bilateral Apr 22 '16 at 15:03

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.

• According to Thom's thesis, only the classes of dimension less than half the dimension of $M$ are representable in general. – Bilateral Apr 15 '16 at 7:51