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Let $F\to M$ denote the vector bundle over embedded Riemann submanifold $M\subset \tilde{M}$.

With each fiber be the set of all bilinear maps $T_pM\times T_pM \to N_pM$.Prove this vector bundle is smooth.

My attempt:To do this we can use Vector bundle chart lemma say for example in Lee's ISM book Prop 10.6.

Which needs to give the local trivialization map $\Phi:\pi^{-1}(U)\to U\times \Bbb{R}^k$,we may choose it as follows:take the adapted orthnormal frame $(E_1,...,E_m,E_{m+1},...,E_n) $ where the first $m$ correspond to the embedded submanifold $M$ that is $(E_1,...,E_m)$ span the tangent space $T_pM$ and $(E_{m+1},...,E_n) $ span $N_pM$.

Then all the bilinear map can be represented as a $\Bbb{R}^{m^2\times(n-m)}$ matrix under the chooice of orthnormal frame.Which gives exactly the local trivialization map.

The key point to check this is a smooth vector bundle is to check that transition map is smooth,then we need to consider two different othnormal frame,and see whether is smooth or not.I find it hard to check this transition is smooth,it's there some idea to handle this problem?

I have alternative idea to handle this problem,Using the natural identification between :linear map $V\to W$ and bilinear form $V\times W^* \to \Bbb{R}$

Hence each fiber is exactly $T_pM\times T_pM \times N^*_pM \to \Bbb{R}$,Then we can do just like what we did in proving mix tensor bundle is smooth.

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It is a general fact that the (functorial) operations of linear algebra extend naturally to vector bundles. You can see this by using the basic result that a vector bundle is determined (up to isomorphism) by a cover $\{U_{\alpha}\}$ of the base space and functions on this cover $\{g_{\alpha\beta}\}:U_{\alpha\beta}\rightarrow\text{GL}(n,\mathbb{R})$ which satisfy the cocyle condition.

For instance, if bundles $E_i$ are determined by cocycles $\{g_{\alpha\beta}^i\}$ for $i=1,2$, you can define their direct sum $E_1\oplus E_2$ as the bundle determined by the transition functions $\{g_{\alpha\beta}^1\oplus g_{\alpha\beta}^2\}$. You indeed have to see differentiability, but that is obvious in the matrix form: $$g_1\oplus g_2=\begin{pmatrix} g_1 & 0\\ 0 & g_2 \end{pmatrix} $$ Similarly, you can define the tensor product and the dual of bundles. Moreover, the identification you mentioned: $$\text{Hom}(V,W)\simeq V^*\otimes W $$ allows the definition of the bundle $\text{Hom}(E_1,E_2)$ as before.

Now your bundle is $\text{Hom}(TM\otimes TM,NM)$.

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  • $\begingroup$ It should be the tensor product and not the direct sum of the tangent bundle with itself. $\endgroup$
    – Deane
    Aug 1, 2021 at 20:20

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