# Tensor product of real line bundles is trivial as a map $\mathbb{R}P^\infty\to\mathbb{R}P^\infty\times\mathbb{R}P^\infty\to\mathbb{R}P^\infty$

The tensor product of a real line bundle with itself is trivial, as is easily seen by looking at the transition functions or checking the Stiefel-Whitney class. Real line bundles are classified by the space $\mathbb{R}P^\infty$, and there is a map $\mathbb{R}P^\infty\times\mathbb{R}P^\infty\overset{\mu}{\to}\mathbb{R}P^\infty$ corresponding to tensor product. Hence we should expect that the composite map $\mathbb{R}P^\infty\overset{\Delta}{\to}\mathbb{R}P^\infty\times\mathbb{R}P^\infty\overset{\mu}{\to}\mathbb{R}P^\infty$ be nullhomotopic, where the first map is the diagonal. Is there a way to see this directly? Is there an explicit description of $\mu$ as a map to projective space?

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(A representative for) $\mu$ is the "infinite Segre embedding"

$$(x_0 : x_1 : \dots) \times (y_0 : y_1 : \dots) \mapsto (x_0 y_0 : x_0 y_1 : x_1 y_0 : \dots)$$

so the composition with the diagonal is the "infinite Veronese embedding"

$$(x_0 : x_1 : \dots) \mapsto (x_0^2 : x_0 x_1 : x_1 x_0 : x_1^2 : \dots).$$

An explicit nullhomotopy from this map to the map with constant value $(1 : 0 : \dots)$ is given by

$$t \times (x_0 : x_1 : \dots) \mapsto \left( x_0^2 + t (1 - x_0^2) : x_0 x_1 (1 - t) : x_1 x_0 (1 - t) : \dots \right).$$

The reason this works is just that we always have $x_0^2 \ge 0$, and moreover at least one of $x_0^2, x_1^2, \dots$ is nonzero. The subspace of $\mathbb{RP}^{\infty}$ where any coordinate is always positive is contractible, being a copy of $\mathbb{R}^{\infty}$, and in this case we can take $x_0^2 + x_1^2 + \dots$ to be the coordinate (I mean the sum of the coordinates with those entries in it; hopefully this is clear). In order for a map to $\mathbb{RP}^{\infty}$ to not be nullhomotopic it must have the property that every coordinate takes the value $0$ at least once.

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This is very cool. Thank you, Qiaochu. The composition of the Segre embedding with the diagonal is sometimes called the Veronese embedding, so here we have the infinite Veronese? –  Joe Hannon Jul 15 at 6:40
Ah yes, that's what I should call it. Thanks! –  Qiaochu Yuan Jul 15 at 6:42
This explanation also shows that the tensor product of complex line bundles is not trivial, which is a nice sanity check. Is there an analogous map for higher rank bundles? Maybe use the Segre embedding composed with the Plücker embedding of the Grassmannian in projective space? –  Joe Hannon Jul 15 at 20:22