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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.

By this question, the etaWe say that a cobordism invariant $\eta$ is not a topological invariant butif 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: AreDetermine whether the other two cobordism invariants mentioned above are topological orand geometric?. In general, how to determine whether aare there any examples of cobordism invariant isinvariants which are geometric but not topological or? 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!

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.

By this question, the eta invariant $\eta$ is not a topological invariant but it is a geometric invariant.

My question: Are the other two cobordism invariants mentioned above topological or geometric? In general, how to determine whether a cobordism invariant is topological or geometric?

Thank you!

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!

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Borromean
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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.

By this question, the eta invariant $\eta$ is not a topological invariant but it is a geometric invariant.

My question: Are the other two cobordism invariants mentioned above topological or geometric? In general, how to determine whether a cobordism invariant is topological or geometric?

Thank you!