# Thom class in homology for defining orientability of vector bundles

A rank $$r$$ real vector bundle $$p : E \to B$$ is said to be orientable if there is a Thom class $$\tau \in H^r (D(E),\partial D(E) ; \mathbb{F})$$ (where $$D(E)$$ is a unit disk bundle and $$\mathbb{F}$$ a field) if its restriction to each fibre $$(D(E)_b,\partial D(E)_b)$$ for $$b \in B$$ is a generator of the group $$H^r ( D(E)_b, \partial D(E)_b ; \mathbb{F})$$.

If we swap the cohomology groups for homology groups (by Poincare duality) does that given an equivalent definition?

Thank you.

• So a generator of $H_n(D(E))$ that is in the image of every induced map $H_n(\partial D(E)_b)\rightarrow H_n(D(E))$, where $b\in B$ and $n=\dim(B)$? This is equivalent, but more difficult to work with. Also it only applies in the special case that $p:E\rightarrow B$ is a smooth Riemannian vector bundle, and even here it is generally easier to work with smooth forms than possibly non-representable homology classes. – Tyrone May 21 at 10:45
• @Tyrone thanks, can you please give me some more detail or a reference about to get from my statement to your statement. Is this Lefschetz duality? I am ok with this working only on a smooth riemannian vector bundle because this is actually what I have. – Nasos Evangelou May 22 at 12:54
• @Tyrone also, the Thom isomorphism in cohomology is $H^k(X,\mathbb{F}) \simeq H^{k+r} (D(E), \partial D(E) ; \mathbb{F})$. Is the correct homology version just $H_k(X,\mathbb{F}) \simeq H_{k+r} (D(E), \partial D(E);\mathbb{F})$ ? Also, can you please tell me where the riemannian structure is needed in what you said? – Nasos Evangelou May 22 at 13:26
• I have just applied Poincare duality, which is induced by taking the cap product with the fundamental class. $D(E)$ is an $(n+r)$-dimensional manifold with boundary, and $B$ is $n$-dimensional. The Thom isomorphism on cohomology is induced by taking the cup product with the Thom class $\tau$. Once you understand how cap and cup products interact you see that the Thom isomorphism in homology is $H_{k+r}(D(E),S(E))\xrightarrow{\cong}H_k(B)$, $x\mapsto \tau\cap x$. – Tyrone May 23 at 9:59
• The maps $H^r(D(E),S(E))\rightarrow H^r(D(E)_b,S(E)_b)$ and $H_{n}(D(E)_b)\rightarrow H_n(D(E))$ are Poincare dual. You don't actually need the Riemannian structure to have a Thom class, but if you are specifically working with the disc and sphere bundles then these need to be defined with a metric. – Tyrone May 23 at 10:27