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Pictures below is from the do Carmo's Riemannian Geometry, I don't know why Gauss Lemma means the boundary of normal ball is a hypersurface, although it is very visualized. Besides, seemly, there is not explicit definition of hypersurface in this book.

Last, the hypersuface orthogonal to the geodesics start from $p$ is also visualized in my view, but I also how to get it. Seemly, it is the process of Gauss lemma what make this point.

enter image description here

enter image description here

Enhao Lan
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  • Gauss Lemma is applied to say that the geodesic spheres are orthogonal to radial geodesics. That these are hypersurfaces is clear, does not need the lemma. – Laz Jul 06 '21 at 01:42
  • @Laz Why it is hypersurfaces is clear ? And what is the strict definition of hypersurfaces, – Enhao Lan Jul 06 '21 at 07:47

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Do Carmo's formulation is ambiguous, and he does not mean that the Gauss Lemma implies that the boundary of a geodesic ball is an hypersurface. The author might have prefered this formulation:

The boundary of a geodesic ball, contained in a normal neighbourhood, is an hypersurface, and by Gauss Lemma, it is normal to geodesic rays starting from $p$.

The fact that small geodesic balls are hypersurfaces comes from the very definition of a normal neighbourhood. Indeed, for $V\subset T_pM$, being a normal neighbourhood of $0 \in T_pM$ means that $\exp_p \colon V \to \exp_p(V) \subset M$ is a diffeomorphism. Thus, as a sphere of radius $\varepsilon$ in $T_pM$ is an hypersurface and for $\varepsilon$ small enough, it is contained in $V$, this imply that its image is a compact hypersurface of $\exp_p(V)$, which is open in $M$. It follows that it is an hypersurface of $M$.

The fact that it is normal to geodesic rays is a direct consequence of Gauss Lemma. If $v \in T_pM$ is a non-zero vector and if $w \perp v$, then, by Gauss Lemma: $$ \forall t, \langle \mathrm{d}\exp_p(tv)v,\mathrm{d}\exp_p(tv)w\rangle = \langle v,w\rangle = 0 $$ and therefore, $\mathrm{d}\exp_p(tv)w$ is orthogonal to $\mathrm{d}\exp_p(tv)v$, which is the tangent vector of $\gamma(t) = \exp_p(tv)$.

Comment the other answer claims a wrong statement. It is false that the exponential map is an immersion. For example, on the unit sphere, if $v$ and $w$ are orthonormal vector at $p \in S^n$, then $\mathrm{d}\exp_p(\pi v) w = 0$ by easy Jacobi fields analysis: indeed $\mathrm{d}\exp_p(tv)(tw) = \sin t W$ where $W$ is the parallel transport of $w$ along $\exp(tv)$. Therefore, $\mathrm{d}\exp_p(\pi v)$ is not injective. However, it is true that in non-positive curvature, it is.

Didier
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Heuristically Gauss' lemma allows us to show that the exponential map is an immersion. Indeed if $v$, $w$ belong to $T_{p}M$, then

$$ \langle d\exp_{p}(v-w),d\exp_{p}(v-w)\rangle = \langle v-w, v-w\rangle= \|v-w\|^{2}.$$

If $d\exp_{p}v=d\exp_{p}w$, then it follows that $\|v-w\|^{2}=0$, and so $v=w$. This proves that $d\exp_{p}$ is an injection, thus $\exp_{p}$ is an immersion.

A sphere in $T_{p}M$ is a smooth submanifold of codimension $1$ in $T_{p}M$. Then the local immersion property of the exponential map implies that it is a local diffeomorphism, so the image of the sphere under the exponential map is still locally a hypersurface, and is thus an immersed submanifold of $M$.

This is the best of my understanding, but I could be wrong. Without much details it is difficult to come up with a correct explanation....

enoughsaid05
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    I am not saying your reasoning cannot be used to prove that the exponential map is an immersion. However, if you got the relation in the third line from applying Gauss Lemma directly to $v = w = v - w$, then the expression beginning the fourth line $d \text{exp}p v = d \text{exp}_p w$, should be read as $d(\text{exp}_p){v-w} (v) = d(\text{exp}p){v-w} (w)$. Don't you see a problem here? How do you suggest to proceed? – Laz Jul 06 '21 at 04:48
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    From what I understand, all your post does is to prove that $d(\text{exp}_p)_u(u)=0$ only if $u=0$. This certainly does not prove that $\text{exp}_p$ is an immersion at $u\in T_p M$. – Laz Jul 06 '21 at 05:02
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    What do you mean by $\mathrm{d}\exp_p(v-w)$? To me, if $v \in T_pM$, then $\mathrm{d}\exp_p(v)$ is a linear map from $T_pM$ to $T_{\exp_p(v)}M$, and is not a vector. Moreover, the exponential map is not an immersion in a general setting: for example, on the sphere, it is not: if $v$ and $w$ are orthonormal at a point, then $\mathrm{d}\exp_p(\pi v) w = 0$, which easily follows from Jacobi fields analysis. – Didier Jul 06 '21 at 08:09