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Let $p:S^3 \to S^2$ be the Hopf fibration which is a result of the standard action of $S^1$ on $S^3$.

Is there a $2$ dimensional vector bundle $\tilde{p}:E \to S^2$ such that $S^3\subset E$ and $\tilde{p}_{|S^3}=p$?

Is there a vector bundle $E$ as above with the following extra condition:The total space $E$ can be acted by $S^1$ with linear isomorphism and this action would be the extension of the standard action of $S^1$ on $S^3$?

Note: One can ask the same question for such extension of an arbitrary principal bundle to an equivariant vector bundle of arbitrary dimension.

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I think both of your questions can by answered positive. Since the Hopf fibration is an $S^1$ bundle it comes as a sphere bundle of complex line bundle over $S^2$. From the Gysin sequence the first Chern class of that line bundle must be a generator of $H^2(S^2;\mathbb Z)$ which determines the isomorphism type of the line bundle.

Moreover a circle action on $S^2$ can be lifted to a line bundle if and only if the first Chern class possess an equivariant extension in $H_{S^1}(S^2;\mathbb Z)$ which here is always true.

Edit: I would like to remark that lifting actions from the base space to a vector bundle is in general a very difficult problem (which is solved for complex line bundles) but unknown (as far as I know) for complex vector bundle of rank higher than 2. There are some results for oriented vector bundles over spheres.

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  • $\begingroup$ But the lifted action from $S^2$ would not extend the action of $S^1$ on $S^3$. $\endgroup$
    – Arkadij
    Commented Jul 29, 2019 at 10:32
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    $\begingroup$ It should. It is an action of $S^1$ on the complex line bundle and you can choose an $S^1$ invariant fiber metric. The sphere bundle of the line bundle to this invariant metric is by construction $S^3$ and again by construction $S^1$ acts on $S^3$. $\endgroup$ Commented Jul 29, 2019 at 10:54

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