# Sphere with given tangent space

Let $$S^n$$ be the unit $$n$$-sphere equipped with its standard metric inherited from $$\mathbb{R}^{n+1}$$, let $$p \in S^n$$, and let $$V \subset T_pS^n$$ be an $$m$$-dimensional subspace of the tangent space at $$p$$, where $$m < n$$. Is it possible to isometrically embed $$S^m$$ into $$S^n$$ so that $$T_pS^m = V$$?

This seems like it should be true, especially if one considers the case $$n=2$$ and $$m=1$$, since given a tangent of $$S^2$$ at $$p$$, it is easy to find a great circle through $$p$$ with the same tangent.

For the general case I was thinking of taking an orthonormal basis of $$V$$, and somehow parametrizing an $$m$$-sphere using it, but this seems like an overkill.

Could someone give me a hint on how to see this?

I think that you can reduce your problem to the case where $$p=e_1$$ and $$V=Vect(e_2,\dots,e_{m+1})$$, where $$e_i=(0,\dots,0,1,0,\dots,0).$$
First $$O_n(\Bbb{R})$$ acts transitively on the sphere so you can assume $$p=e_{1}$$. If you take $$(a_1,\dots,a_k)$$ an orthogonal basis for $$V\subset e_1^\perp$$, there is an orthogonal transformation taking each $$a_i$$ to $$e_{i+1}$$ and fixing $$e_1$$, so you can assume $$V=Vect(e_2,\dots,e_{m+1})$$.
Finally you can take $$i:(x_1,\dots,x_{m+1})\in S^m\longmapsto (x_1,\dots x_{m+1},0,\dots,0)\in S^n$$