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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?

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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 $$i:(x_1,\dots,x_{m+1})\in S^m\longmapsto (x_1,\dots x_{m+1},0,\dots,0)\in S^n$$

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    $\begingroup$ That seems like exactly the argument I was after, thanks! $\endgroup$ – MisterRiemann May 6 at 15:46

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