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(Edited after taking into account Tom Goodwillie's answer.)

Let $E \rightarrow X$ be an orientable vector bundle.

In this MO answer it is explained how to obtain a representative of the Thom class (call this a Thom form) which pulls back to the natural Euler-form given by Chern-Weil theory. This form depends on the choice of: a metric and a compatible connection on $E$, and a cutoff function $\rho:[0,\infty)\rightarrow [0,\infty)$ which is equal to $−1$ near $0$ and equal to $0$ on $[1,\infty)$.

If we are given a Thom form for $i^*E \rightarrow \partial X$ constructed via the above method, one can extend the metric and connection to $E \rightarrow X$ using a partition of unity argument, and use the above method to obtain a Thom form for $E \rightarrow X$ extending the Thom form on the boundary. However, it is not clear to me whether every choice of Thom form can be constructed using this method.

As answered by Tom Goodwillie in the comments, if $i: \partial X \hookrightarrow X$ is a smooth manifold with boundary and $E \rightarrow X$ an orientable vector bundle then any Thom form for $i^*E \rightarrow \partial X$ can be extended to a Thom form for $E \rightarrow X$. But, this extension might not arise from a metric and compatible connection.

If this general question proves too hard, I am trying to use it to obtain an answer to the following more specific situation: suppose that $i^*E = E_1 \oplus E_2 \rightarrow \partial X$, then one can choose metrics and compatible connections $g_i$ and $\nabla_i$ for $E_i$, and a function $\rho$ as before, and use this to obtain Thom forms $\tau_i \in \Omega^{rk(E_i)}_{cv}(E_i)$, then $\tau_1 \wedge \tau_2$ is a Thom form for $i^*E \rightarrow \partial X$. Are there choices of metric and connection which give rise to such an extension? One might hope to prove this directly from the formula for the Thom form, but it is not clear to me what happens in the construction of the global angular form of a direct sum.

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    $\begingroup$ For the question as stated in the first sentence, the answer is "yes". (But maybe you really want to use a metric and a connection.) If $\sigma$ represents the Thom class of $E$ and $\tau$ represents the Thom class of $i^\ast E$, then the difference between $\tau$ and the restriction of $\sigma$ is $d\rho$ for some form $\rho$ (with the appropriate kind of support), so extend $\rho $ to $E$ and add the differential of that to $\sigma$ to get a Thom form that restricts to $\tau$. $\endgroup$ Commented Apr 15 at 10:42
  • $\begingroup$ Thanks @TomGoodwillie, for some reason I had convinced myself that partition of unity arguments wouldn't work, as we are trying to obtain a closed form. However, what you write works! The second part of the question is indeed different from the first question. For my purposes your answer is sufficient, but the second part is hopefully of independent interest. I will edit the question to make this clear. $\endgroup$ Commented Apr 15 at 11:06

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