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If $\gamma \colon S^1 \rightarrow SO(2)$, we define $E_{\gamma}= D_{a}^2 \times \mathbb{R}^2 \sqcup D_{b}^2 \times \mathbb{R}^2 / \sim$, where $(x,v) \sim (x,\gamma(x)\cdot v)$ for $(x,v)\in S_{a}^1 \times \mathbb{R}^2$. I need to prove that it is an oriented vector bundle, and compute the Euler class in terms of $\gamma$.

I have defined $\pi \colon E_{\gamma} \rightarrow S^2$ to be $\pi([(x,v])=[x]$, if we take $S^2= D_{a}^2 \sqcup D_{b}^2 / (S_{a}^1 \sim S_{b}^1)$.

I have proved that it is a vector bundle of dimension 2, but I have troubles when computing the Thom class and the Euler class. First of all, I have to define a Riemannian metric on the bundle.

I have defined it in this way: $g\colon E_{\gamma} \times_{\pi} E_{\gamma} \rightarrow \mathbb{R}$ such that $g([(x_{1},v_{1})],[(x_{1},v_{2}])=\langle v_{1},v_{2} \rangle$.

To compute $D(E_{\gamma})= \{[(x,v)]\in E_{\gamma} \mid g([(x,v)],[(x,v)])\leq 1 \}$, I have thought that it is $\text{int}(D_{a}^2)\times D^2 \cup \text{int}(D_{b}^2)\times D^2 \cup S^1 \times D^2$, but that is not possible since $H^2(D(E_{\gamma}))$ must be $\mathbb{Z}$.

Can anyone help me with the Thom and Euler class, please? Thank you.

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1 Answer 1

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Your definition of a metric should have $x_1$ in both arguments, not just one.

Why do you think it's not well-defined on $S^1$? Perhaps because you might get either $\langle v_1, v_2 \rangle$ or $\langle \gamma(x) v_1, \gamma(x) v_2 \rangle$? Remember, $\gamma(x) \in SO(2)$, so it preserves inner products. That means that these two potentially contradictory dot-products must actually always be the same. (If you only knew that the image of $\gamma$ was in $GL(2)$, then you could be in trouble! You might want to think about what you could say if it was guaranteed to be in $O(2)$ rather than $SO(2)$. )

It seems to me that you're headed in the right direction.

post comment addition

You write "However $[(x, v_1)] = [ (x, \gamma(x) \cdot v_1)]$", but that's misleading. The first $x$ is in $D_a$, while the second is in $D_b$.

If you write points in $D^2_a$ in the form $(x, a)$ (i.e., you "tag" them with a second term), then the right version of your statement is that $$ [ ( (x, a), v ] \sim [ ((x, b), \gamma( (x, a) ) \cdot v_1 ]. $$

Now my definition of the dot product of two classes $p$ and $q$ whose basepoints happen to be on the boundary is this: for equivalence class $p$, pick a representative of the form $ (x, a), v$; for $q$, pick a representative of the form $ (x,a), w$.

The dot product is then defined to be $\langle v, w \rangle$. Alternatively, if you choose representatives $(x, b), v'$ and $(x, b), w'$, then the dot product is $\langle v', w' \rangle$. And if you choose $(x, a), v$ and $(x, b), w$, then the dot product would be computed as $$ \langle \gamma(x,a)\cdot v, w \rangle>. $$

You should check that all three of these alternatives (and the missing 4th alternative) produce the same results.

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  • $\begingroup$ But imagine that $x\in S_{a}^1$. Then, $g([(x,v_{1})],[(x,v_{2})])=\langle v_{1}, v_{2} \rangle$. However, $[(x,v_{1})]=[(x,\gamma(x)\cdot v_{1})]$, and in this case I have $\langle \gamma(x)\cdot v_{1},v_{2}\rangle$, and those do not have to be equal. I don't know which part I am understanding wrong. $\endgroup$
    – Laura
    Commented Jan 13, 2018 at 15:13
  • $\begingroup$ See post-comment addition. $\endgroup$ Commented Jan 13, 2018 at 15:34
  • $\begingroup$ To compute $D(E_{\gamma})=\{[(x,v)]\in E_{\gamma} \mid g([(x,v)],[(x,v)])\leq 1 \}$, wouldn't that be $\text{int}(D_{a}^2)\times D^2 \cup \text{int}(D_{b}^2)\times D^2 \cup S^1 \times D^2$? and $S(E)$ the same but with $S^1$ instead of $D^2$? I believe that no, because $H^2(D(E_{\gamma}))$ should be $\mathbb{Z}$, but I do not understand why. $\endgroup$
    – Laura
    Commented Jan 13, 2018 at 17:46
  • $\begingroup$ Yes and no -- -"would that be ... (union of three things) "; the answer here is "yes", but it's more than a set of points -- there's a topology, too. You have to say how the boundary of $D^2_a$ (cross $D^2$ or $S^1$ joins up with the boundary of $D^2_b$ (cross $D^2$ or $S^1$). Probably better to think of it as $D^2_a \times D^2$ an $D^2_b \times D^2$, with an identification, namely, $(x, a), v \leftrightarrow (x, b), \gamma(x,a)\cdot v$. $\endgroup$ Commented Jan 13, 2018 at 18:36
  • $\begingroup$ Ok! I understand the difference, but with that definition I do not have any intuition to compute the cohomology of the pair $(D(E_{\gamma}),S(E_{\gamma}))$, so I do not know how to compute the Thom and Euler class... $\endgroup$
    – Laura
    Commented Jan 13, 2018 at 19:02

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