# Cobordism invariants: topological v.s. geometric

Some cobordism invariants are not cohomology classes. Such as the $$\mathbb{Z}_{16}$$-valued eta invariant $$\eta$$ of $$\Omega_4^{Pin^+}$$, the $$\mathbb{Z}_8$$-valued Arf-Brown-Kervaire invariant ABK of $$\Omega_2^{Pin^-}$$, and the $$\mathbb{Z}_4$$-valued quadratic enhancement $$q(a)$$ of $$\Omega_2^{Pin^-}(B\mathbb{Z}_2)$$ where $$a\in H^1(M,\mathbb{Z}_2)$$. The last invariant is defined as follows: Any 2-manifold $$M$$ always admits a $$Pin^-$$ structure. $$Pin^-$$ structures are in one-to-one correspondence with quadratic enhancements $$q: H^1(M,\mathbb{Z}_2)\to\mathbb{Z}_4$$ such that $$q(x+y)-q(x)-q(y)=2\int_M x\cup y\mod4.$$ In particular, $$q(x)=\int_M x\cup x\mod2.$$ There are many more such examples, I only mention these.

We say that a cobordism invariant is topological if it can be defined purely using topological data, for example, if it is a cohomology class. While we say that a cobordism invariant is geometric if it can be defined purely using geometric data like metric, connection, and curvature.

These two definitions have no confliction, a cobordism invariant can be both topological and geometric.

My question: Determine whether the cobordism invariants mentioned above are topological and geometric. In general, are there any examples of cobordism invariants which are geometric but not topological? Are there any examples of cobordism invariants which are topological but not geometric?

For example, the eta invariant $$\eta$$ is discussed in this question.

Thank you!

• Can you give precise definitions to "a cobordism invariant is topological", resp "geometric"? To me some invariant associated to a manifold with extra structure is topological if it depends on little more than the smooth structure, eg an invariant of spin manifolds might be called topological because there are finitely many spin structures on a given smooth manifold. That the $\eta$ invariant of index theory normally depends on structure like the metric/connection just says to me that your $\Bbb Z/16$-valued $\eta$ extracts topological information from a normally geometric object. – Mike Miller Mar 25 at 18:18
• @MikeMiller Thank you! I have added the definitions. – Borromean Mar 26 at 4:29

$$\newcommand{\ko}{\mathit{ko}} \newcommand{\MTSpin}{\mathit{MTSpin}} \newcommand{\Z}{\mathbb Z}$$ As Mike points out in his comment, it's not obvious what it means for a bordism invariant is “topological” or “geometric.” The bordism invariants you mention can be described topologically, and some bordism invariants which you might think of as topological also admit geometric descriptions.

Pontrjagin numbers are oriented bordism invariants which look topological: take a cohomology class on your manifold and evaluate it on the fundamental class. For example, there's an injective map $$\Omega_4^{\mathrm{SO}}\to\mathbb Z$$ sending a closed, oriented $$4$$-manifold $$X$$ to $$\langle p_1(X), [X]\rangle$$; here $$p_1$$ is the first Pontrjagin class of $$X$$.

However, there is an equivalent, “geometric” definition of this invariant: choose a connection on the vector bundle $$TX\to X$$, and let $$F$$ be its curvature. Then one can make sense of $$\mathrm{tr}(F\wedge F)\in\Omega^4(X)$$, and Chern-Weil theory proves that $$-\frac{1}{8\pi^2}\int_X \mathrm{tr}(F\wedge F) = \langle p_1(X), [X]\rangle.$$ Certainly a connection is geometric data, so this invariant is both “topological” and “geometric.”

The pin$$^\pm$$ bordism invariants you mention admit geometric interpretations (via $$\eta$$-invariants), but also topological ones, though the topology is harder to see. First, let's reframe the above bordism invariant in terms of Thom spectra: the characteristic class $$p_1\in H^4(B\mathrm{SO})$$ defines via the Thom isomorphism a cohomology class in $$\tilde H^4(M\mathrm{SO})$$, hence a map $$M\mathrm{SO}\to\Sigma^4 H\mathbb Z$$, and upon taking $$\pi_4$$, we obtain the map $$\Omega_4^{\mathrm{SO}}\to\mathbb Z$$ from above.

The point of this reformulation is that by replacing $$H\mathbb Z$$ (i.e. ordinary cohomology) with other cohomology theories, we can describe the invariants you've mentioned above.

• As a warm-up, take the Arf invariant of a spin surface, which defines an isomorphism $$\Omega_2^{\mathrm{Spin}}\to\mathbb Z/2$$. There are several different ways to define it, but here's one in line with our above description of $$p_1$$: we have the Atiyah-Bott-Shapiro map $$\mathit{ABS}\colon\MTSpin\to\ko$$,1 and upon taking $$\pi_2$$, this yields a map $$\Omega_2^{\mathrm{Spin}}\to \pi_2\ko\cong\mathbb Z/2$$. One can unwind this definition through the Pontrjagin-Thom isomorphism and obtain a description of the Arf invariant through integration of $$\ko$$-cohomology classes.
• Next, the Arf-Brown-Kervaire invariant. There is a splitting $$\mathit{MTPin}^-\simeq\MTSpin\wedge \Sigma^{-1}\mathit{MO}_1$$, where $$\mathit{MO}_1$$ is the Thom spectrum of the tautological bundle $$\sigma\to B\mathrm O_1$$. Therefore, smashing the Atiyah-Bott-Shapiro map with $$\Sigma^{-1}\mathit{MO}_1$$, we obtain a map $$\mathit{MTPin}^-\simeq\MTSpin\wedge \Sigma^{-1}\mathit{MO}_1\longrightarrow \ko\wedge\Sigma^{-1}\mathit{MO}_1.$$ Taking $$\pi_2$$, we get a map $$\Omega_2^{\mathrm{Pin}^-}\to \pi_2(\ko\wedge \Sigma^{-1}\mathit{MO}_1)\cong \widetilde{\ko}_3(\mathit{MO}_1)\cong\Z/8,$$ and this is the Arf-Brown-Kervaire invariant.2 This can also be described in terms of a pushforward in twisted $$\ko$$-theory.
• The same approach works $$\Omega_4^{\mathrm{Pin}^+}\to\Z/16$$, using this time the splitting $$\mathit{MTPin}^+\simeq\MTSpin\wedge\Sigma\mathit{MTO}_1$$; here $$\mathit{MTO}_1$$ is the Thom spectrum of the virtual bundle $$-\sigma\to B\mathrm O_1$$. Smashing the Atiyah-Bott-Shapiro map with $$\Sigma\mathit{MTO}_1$$ and taking $$\pi_4$$ yields a map $$\Omega_4^{\mathrm{Pin}^+}\to \widetilde{\ko}_3(\mathit{MTO}_1)\cong\Z/16$$.
• The same approach works for $$\Omega_2^{\mathrm{Pin}^-}(B\Z/2)\to\Z/4$$: $$\Omega_2^{\mathrm{Pin}^-}(B\Z/2)\cong\pi_2(\MTSpin\wedge \Sigma^{-1}\mathit{MO}_1\wedge (B\Z/2)_+),$$ which maps to $$\pi_2(\ko\wedge \Sigma^{-1}\mathit{MO}_1\wedge (B\Z/2)_+) = \widetilde{\ko}_3(\mathit{MO}_1\wedge (B\Z/2)_+)\cong\Z/4.$$

If you're asking which bordism invariants come from cohomology classes, the answer is that characteristic classes for $$n$$-dimensional $$G$$-bordism live in $$H^n(BG;A)$$, where $$A$$ is a coefficient group; a $$G$$-structure on a manifold $$M$$ pulls back this class to $$M$$, and then we evaluate it on the fundamental class. In general, this won't capture everything: for example, if two $$G$$-structures have the same underlying orientation, their values on any cohomological bordism invariant will agree. Hence, for example, the Arf invariant isn't cohomological, as there are different spin structures on a torus which induce the same orientation, but have different Arf invariants. This is why topological descriptions of such bordism invariants use generalized cohomology.

1: Here, given a $$G$$-structure $$X\to B\mathrm O$$, $$\mathit{MTG}$$ means the Thom spectrum whose homotopy groups are the bordism groups of manifolds with a $$G$$-structure on the tangent bundle, rather than on the stable normal bundle. This distinction is irrelevant for spin bordism, but important for pin$$^\pm$$ bordism.

2: The Arf-Brown-Kervaire invariant depends on a choice of an 8th root of unity; depending on this choice, we might need to compose with an automorphism of $$\Z/8$$ to obtain “the” Arf-Brown-Kervaire invariant. The same caveat applies to the $$\Z/16$$ and $$\Z/4$$ invariants.

• Thank you for your nice answer! I have added the definitions for "topological" and "geometric" and I ask whether they contain each other. – Borromean Mar 26 at 4:39