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Let's suppose $M$ and $N$ are manifolds.

Here 1 one states that \begin{equation} T(M \times N) \cong \pi_{M}^{*}(T M) \times \pi_{N}^{*}(T N) \end{equation} But in my notes I find \begin{equation} T(M \times N) \cong (T M) \times(T N) \end{equation} and in this question 2 I find the equation \begin{equation} T(M \times N) \cong\pi_{M}^{*} (T M) \oplus \pi_{N}^{*} (T N) \end{equation}

Now I have got two questions:

  1. Which of the equations above is right (or are all of them right)?
  2. What exactly ist the pullback of the projection map? Is it just the invers of the projection map?
NicAG
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  • Yes, they are all the same. The pullbacks of the projection maps are implicit in the first and third case. – Berci Apr 14 '20 at 15:14

1 Answers1

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Given a smooth map $f:X\to Y$ of manifolds and a smooth vector bundle $p:E\to Y$, the pullback vector bundle is defined as $$f^*E=\{(x,e)\in X\times E:f(x)=p(e)\},$$ which is a submanifold of $X\times E$ and maps to $X$ by the first projection. The vector space structure on the fiber $(f^*E)_x$ over $x\in X$ is defined by observing that the second projection gives a bijection $(f^*E)_x\to E_{f(x)}$, and so you pull back the vector space structure on $E_{f(x)}$ along this bijection. So, you can think of $f^*E$ as a vector bundle over $X$ such that the fiber over a point $x\in X$ is given by the fiber of $E$ over $f(x)$.

Now $TM$ is a vector bundle on $M$ and $\pi_M:M\times N\to M$, so the pullback $\pi_M^*(TM)$ is a vector bundle on $M\times N$. Similarly, $\pi_N^*(TN)$ is also a vector bundle on $M\times N$. We can then form the fiberwise direct sum (or equivalently, direct product) of these two vector bundles to get another vector bundle $\pi_M^*(TM)\oplus \pi_N^*(TN)=\pi_M^*(TM)\times \pi_N^*(TN)$ on $M\times N$, which your first and third statements correctly say is isomorphic to $T(M\times N)$.

The second statement is also correct, but it has a different meaning. It is just talking about the tangent spaces as manifolds, not as vector bundles: it says that the manifold $T(M\times N)$ is diffeomorphic to the product of the manifolds $TM$ and $TN$. So the $\times$ symbol in $TM\times TN$ has a different meaning than in the first statement: it denotes the ordinary Cartesian product of manifolds, whereas in the first statement it denotes the fiberwise Cartesian product of vector bundles.

Eric Wofsey
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  • Thanks for the answer. Could we view $\pi^_M(TM)$ as $TM \times {0} \subset M \times N$ ( $\times$ is the cartesian product of vector bundles). And is $\pi^_M(TM) \cong TM$ – NicAG Apr 14 '20 at 15:38
  • I don't know what $TM \times {0} \subset M \times N$ means; maybe you meant to write something else? And no, $\pi^_M(TM) \not\cong TM$. $\pi^_M(TM)$ is a vector bundle on $M\times N$ and $TM$ is a vector bundle on $M$. – Eric Wofsey Apr 14 '20 at 15:51
  • Let $(\pi^M ,M \times N,p)$ be the pullback bundle. Then $p( \pi{M}^{} (TM)) =M \times {0}$. Isn't it possible to restrict $(\pi^M ,M \times N,p)$ to $M \times {0}$, so that $ \pi{M}^{}(TM) \cong TM $ – NicAG Apr 14 '20 at 16:02
  • What is the map $p$? How is $p( \pi_{M}^{*} (TM)) =M \times {0}$? – Eric Wofsey Apr 14 '20 at 16:08
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    It is true that if you pick a point $n\in N$ and restrict the bundle $\pi^*_M(TM)$ to $M\times {n}\subseteq M\times N$, you get a bundle on $M\times {n}$ that is isomorphic to $TM$ if you identify $M\times{n}$ with $M$. – Eric Wofsey Apr 14 '20 at 16:10
  • $p$ ist the projection of the vector bundle $\left(\pi_{M}^{}(TM), M \times N, p\right)$. Let $(M,TM,p')$ be the Tangent bundle to M, then $\pi_{M}^{}(TM)={(b=(n,m),v) \in (M \times N) \times TM : \pi_M(m,n)=m=p'(v) }, v \in TM, b=(m,n) \in M \times N$. – NicAG Apr 14 '20 at 16:16
  • Ok so $p(\pi_M^* (TM))$ is not $M \times {0}$. Anyway thank you. – NicAG Apr 14 '20 at 16:19