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

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.

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.

• I just cannot read this. But how did you end up with $TS^m$ mapping to $T\Bbb R^{2m+2}$ rather than $T\Bbb R^{m+1}$? Jun 7, 2021 at 22:06
• @TedShifrin This is indeed an error : fixed Jun 7, 2021 at 22:14
• @TedShifrin Modulo identifications, is it correct that $T_{(p, \xi)}(TS^m) = (p, \xi)^\vdash$ using the definition of the last set as in the answer? Jun 7, 2021 at 22:19
• Shouldn't that be a $2m$-dimensional subspace of $\Bbb R^{2(m+1)}$? Jun 7, 2021 at 22:28
• Yes, I agree that this is correct. And it is in fact obvious that the tangent space to the fiber is constant along the fiber, since the second equation is independent of $y$ when $q=0$. Jun 8, 2021 at 18:22

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.