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Let $X$ and $Y$ be connected smooth manifolds. Let $L$ be a topological real line bundle over $X\times Y$. Then we know that the isomorphism class of such a line bundle is determined by its first Stiefel-Whitney class $w_1(L)\in H^1(X\times Y,\mathbf{Z}/2\mathbf{Z})$.

I would like to have an example of a nontrivial line bundle $L$ (so $w_1(L)\neq 0$), such that

(1) $\forall y\in Y$ we have $w_1(L)|_{X\times\{y\}}=0\in H^1(X\times\{y\},\mathbf{Z}/2\mathbf{Z})$.

(2) $L$ is not the pullback of a line bundle over $Y$.

I don't see how to construct such a line bundle. May be there is some trick using Kunneth's formula.

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    $\begingroup$ This holds trivially if $L$ is the pullback of a nontrivial line bundle on $Y$. Do you want your condition to be true simultaneously for both $X$ and $Y$? $\endgroup$ Mar 27, 2013 at 19:16
  • $\begingroup$ Yes, I forgot to mention that I don't want $L$ to be coming from the pullback of a line bundle over $Y$. $\endgroup$ Mar 27, 2013 at 21:23

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Denote by $p_X$ the natural projection $X\times Y\to X$ and define $p_Y$ similarly. Kunneth formula shows that any $\newcommand{\bZ}{\mathbb{Z}}$ $w\in H^1(X\times Y,\bZ/2)$ has the form

$$ w= p_X^* u+p_Y^* v,\;\;u\in H^1(X,\bZ/2),\;\;v\in H^1(Y,\bZ/2). $$

This proves that any real line bundle $L\to X\times Y$ has the form

$$ L= p_X^*L_X\otimes p_Y^*L_Y, $$

with

$$w_1(L)=p_X^*w_1(L_X)+p^*_Y w_1(L_Y). $$

Moreover

$$ L|_{X\times y}= L_X. $$

The line bundles with the properties you want are all pullbacks of nontrivial line bundles on $Y$.

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    $\begingroup$ I see, so this explains why I was not able to construct such an example. What about complex line bundle? $\endgroup$ Mar 27, 2013 at 21:38
  • $\begingroup$ Ok, I was being idiot $\endgroup$ Mar 27, 2013 at 22:20
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    $\begingroup$ Complex line bundles are classified by $H^2$ (with coefficients in $\mathbb{Z}$), so you can get extra line bundles on a product that come from $H^1(X)\otimes H^1(Y)$. The simplest example of this is for $X=Y=S^1$: every complex line bundle on $X$ or $Y$ is trivial, but $H^2(X\times Y)=\mathbb{Z}$. $\endgroup$ Mar 27, 2013 at 23:32

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