Proof verification of the tangent space of the tangent bundle of the sphere using the canonical identifications

Question

Let $\iota : S^m \hookrightarrow \mathbb{R}^{m+1} $ and $\iota_T : TS^m \hookrightarrow TR^{m+1}$ be embeddings and let $J_v : V \xrightarrow{\cong} T_vV$ be the canonical isomorphism where $V$ is a vector space. Then one can show that $$ D\iota(p)(T_pS^m) = J_{\iota(p)}(\iota(p)^\bot) $$ where $x^\bot = \{y \in \mathbb{R}^{m+1} | \langle x, y\rangle\ = 0\}$. I would like to continue to use these maps for the moment to understand all the details and different objects before progressively get rid of these maps for clarity in a second time.

• Is this similar description for the tangent space of the tangent bundle correct?

Recall that the tangent bundle is $$TS^m = \left\{(p, \xi) \in S^m \times T_pS^m | \left\langle\iota(p), J_{\iota(p)}^{-1} \circ D\iota(p)(\xi)\right\rangle = 0\right\}. $$ Given $(p, \xi) \in TS^m$, consider a smooth curve $\gamma : (-\varepsilon,\varepsilon) \rightarrow TS^m$ such that $$ \gamma(0) = (p, \xi), \quad \dot{\gamma}(0) = (q, \zeta)$$ where $(q, \zeta) \in T_{(p, \xi)}(TS^m).$ Then by differentiating at time $t=0$ the fact that $\mathrm{pr}_1 \circ \gamma$ has image in $S^m$, we get $$ 0 = \frac{d}{dt}\bigg|_{t=0} (|\iota \circ \mathrm{pr}_1 \circ \gamma(t)|^2) = \left\langle \iota(p), J_{\iota(p)}^{-1} \circ D\iota(p)(q)\right\rangle,$$ and by differentiating at time $t = 0$ the the scalar product condition of $TS^m$, we get \begin{align} 0 &= \frac{d}{dt}\bigg|_{t=0} \left(\left\langle \iota \circ pr_1 \circ \gamma(t), J_{\iota \circ \mathrm{pr}_1 \circ \gamma(t)}^{-1} \circ D\iota(\mathrm{pr}_1 \circ \gamma(t))(\mathrm{pr_2} \circ \gamma(t)) \right\rangle\right) \\ &= \left\langle \iota(p), J_{(\iota(p), J_{\iota(p)}^{-1} \circ D\iota(p)(\xi))}^{-1} \circ D\iota_T(p, \xi)(\zeta)\right\rangle + \left\langle J_{\iota(p)}^{-1} \circ D\iota(p)(\xi), J_{\iota(p)}^{-1} \circ D\iota(p)(q)\right\rangle. \end{align} Since $(q, \zeta)$ was arbitrary, we have the inclusion $$ D\iota_T(p, \xi)(T_{(p, \xi)}(TS^m)) \subset J_{(\iota(p), J_{\iota(p)}^{-1} \circ D\iota(p)(\xi))}((\iota(p), J_{\iota(p)}^{-1} \circ D\iota(p)(\xi))^\vdash) $$ where $(x, y)^\vdash = \{(q, \zeta) \in \mathbb{R}^{m+1} \times \mathbb{R}^{m+1} | \langle x, q\rangle = 0 = \langle x, \zeta \rangle + \langle y, q\rangle\}.$ Since, both of the vector spaces are of dimension $2m$, we have equality.

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Edits

To be more readable, this is what the computation would look like if we use the identifications : We recall that $$ TS^m = \{ (p, \xi) \in S^m \times \mathbb{R}^{m+1} | \langle p, \xi \rangle = 0 \}. $$ Given $(p, \xi) \in TS^m$, consider a smooth curve $\gamma : (-\varepsilon,\varepsilon) \rightarrow TS^m : t \mapsto (p(t), \xi(t))$ such that $$ \gamma(0) = (p, \xi), \quad \dot{\gamma}(0) = (q, \zeta)$$ where $(q, \zeta) \in T_{(p, \xi)}(TS^m).$ Then by differentiating at time $t=0$ the fact that $|p(t)|^2=1$, we get $$ 0 = \langle p, q \rangle, $$ and by differentiating at time $t = 0$ the scalar product condition on $TS^m$, we get $$ 0 = \langle p, \zeta \rangle + \langle \xi, q \rangle. $$ Since $(q, \zeta)$ was arbitrary, we have the inclusion $$T_{(p,\xi)}(TS^m) \subset (p, \xi)^\vdash,$$ and we conclude by dimension.

Answer 1

I didn't fully read your first computations, but your second one is right. In general, we have the canonical identification $T\Bbb{R}^m\cong \Bbb{R}^m\times \Bbb{R}^m$ given by $[\gamma]\mapsto (\gamma(0),\gamma'(0))$. In words, we take an equivalence class of curves and we keep track of the base point and the vector-part (the derivative is possible since $\gamma$ takes values in the vector space $\Bbb{R}^m$).

Now, a slight generalization of what you've done is the following. Let $f:\Bbb{R}^n\to\Bbb{R}^k$ be a smooth submersion and $\rho\in\Bbb{R}^k$ a regular value for $f$. Then, as we know from the inverse/implicit function/regular-value theorem, $S:= f^{-1}(\{\rho\})$ is a smooth $n-k$-dimensional embedded submanifold of $\Bbb{R}^n$, and the tangent space at a point $p\in S$ is given by (after the canonical identification) $\ker Df_p$. Now, the tangent bundle can be identified as \begin{align} TS=\{(p,\xi)\in S\times \Bbb{R}^n\,|\,\,Df_p(\xi)=0\} \end{align} Now, the tangent space $T_{(p,\xi)}(TS)$ can be calculated exactly as you have just done. Take a curve $t\mapsto \gamma(t)=(p(t),\xi(t))$ in $TS$ passing through $(p,\xi)$ when $t=0$. Then, $t\mapsto p(t)$ is a curve in $S$, so $Df_p[p'(0)]=0$. Next, we also have the condition that for all $t$, $Df_{p(t)}[\xi(t)]=0$. Differentiating this and setting $t=0$, we get (by "product" and chain rule) \begin{align} D^2f_{p(0)}[p'(0),\xi(0)] +Df_{p(0)}[\xi'(0)]&=0. \end{align} Once again, arguing using dimension reasons (both are $2(n-k)$-dimensional), we find that \begin{align} T_{(p,\xi)}(TS)&=\{(q,\zeta)\in\Bbb{R}^n\times\Bbb{R}^n\,|\,\,Df_p(q)=D^2f_p[q,\xi]+Df_p[\zeta]=0\} \end{align}

In the case of spheres, we have $f(x)=\langle x,x \rangle$, so $Df_p(\xi)=2\langle p,\xi \rangle$ and $D^2f_p[v,w]=2\langle v,w\rangle$, which coincides exactly with what you found.