Gauss's lemma (Riemannian geometry)

Gauss's lemma (Riemannian geometry)

In Riemannian geometry, Gauss's lemma asserts that any sufficiently small sphere centered at a point in a Riemannian manifold is perpendicular to every geodesic through the point. More formally, let "M" be a Riemannian manifold, equipped with its Levi-Civita connection, and "p" a point of "M". The exponential map is a mapping from the tangent space at "p" to "M"::mathrm{exp} : T_pM o Mwhich is a diffeomorphism in a neighborhood of zero. Gauss' lemma asserts that the image of a sphere of sufficiently small radius in "T"p"M" under the exponential map is perpendicular to all geodesics originating at "p". The lemma allows the exponential map to be understood as a radial isometry, and is of fundamental importance in the study of geodesic convexity and normal coordinates.


We define on M the exponential map at pin M by:exp_p:T_pMsupset B_{epsilon}(0) longrightarrow M,qquad vlongmapsto gamma(1, p, v),where we have had to restrict the domain T_pM by definition of a ball B_epsilon(0) of radius epsilon>0 and centre 0 to ensure that exp_p is well-defined, and where gamma(1,p,v) is the point qin M reached by following the unique geodesic gamma passing through the point pin M with tangent frac{v}{vert vvert}in T_pM for a distance vert vvert . It is easy to see that exp_p is a local diffeomorphism around 0in B_epsilon(0). Let alpha : I ightarrow T_pM be a curve differentiable in T_pM such that alpha(0):=0 and alpha'(0):=v . Since T_pMcong mathbb R^n, it is clear that we can choose alpha(t):=vt . In this case, by the definition of the differential of the exponential in 0 applied over v , we obtain:

:T_0exp_p(v) = frac{mathrm d}{mathrm d t} Bigl(exp_pcircalpha(t)Bigr)Bigvert_{t=0} = frac{mathrm d}{mathrm d t} Bigl(exp_p(vt)Bigr)Bigvert_{t=0}=frac{mathrm d}{mathrm d t} Bigl(gamma(1,p,vt)Bigr)Bigvert_{t=0}= gamma'(t,p,v)Bigvert_{t=0}=v.

The fact that exp_p is a local diffeomorphism and that T_0exp_p(v)=v for all vin B_epsilon(0) allows us to state that exp_p is a local isometry around 0 , i.e.

:langle T_0exp_p(v), T_0exp_p(w) angle_0 = langle v, w angle_pqquadforall v,win B_epsilon(0).

This means in particular that it is possible to identify the ball B_epsilon(0)subset T_pM with a small neighbourhood around pin M. We can see that exp_p is a local isometry, but we would like it to be rather more than that. We assert that it is in fact possible to show that this map is a radial isometry !

The exponential map is a radial isometry

Let pin M. In what follows, we make the identification T_vT_pMcong T_pMcong mathbb R^n.Gauss's Lemma states:

Let v,win B_epsilon(0)subset T_vT_pMcong T_pM and M i q:=exp_p(v). Then, :langle T_vexp_p(v), T_vexp_p(w) angle_v = langle v,w angle_q.

For pin M, this lemma means that exp_p is a radial isometry in the following sense: let vin B_epsilon(0), i.e. such that exp_p is well defined. Moreover, let q:=exp_p(v)in M. Then the exponential exp_p remains an isometry in q , and, more generally, all along the geodesic gamma (in so far as gamma(1,p,v)=exp_p(v) is well defined)! Then, radially, in all the directions permitted by the domain of definition of exp_p , it remains an isometry.


Recall that

:T_vexp_p : T_pMcong T_vT_pMsupset T_vB_epsilon(0)longrightarrow T_{exp_p(v)}M.

We proceed in three steps:
* "T_vexp_p(v)=v " : let us construct a curve alpha : mathbb R supset I ightarrow T_pM such that alpha(0):=vin T_pM and alpha'(0):=vin T_vT_pMcong T_pM. Since T_vT_pMcong T_pMcong mathbb R^n, we can put alpha(t):=v(t+1) . We find that, thanks to the identification we have made, and since we are only taking equivalence classes of curves, it is possible to choose alpha(t) = vt (these are exactly the same curves, but shifted (###décalées###), because of the domain of definition I ; however, the identification allows us to gather them (###ramener###) around 0 !!!). Hence,

:T_vexp_p(v) = frac{mathrm d}{mathrm d t}Bigl(exp_pcircalpha(t)Bigr)Bigvert_{t=0}=frac{mathrm d}{mathrm d t}gamma(t,p,v)Bigvert_{t=0} = v.

Now let us calculate the scalar product langle T_vexp_p(v), T_vexp_p(w) angle.

We separate w into a component w_T tangent to v and a component w_N normal to v . In particular, we put w_T:=alpha v , alphain mathbb R.

The preceding step implies directly:

:langle T_vexp_p(v), T_vexp_p(w) angle = langle T_vexp_p(v), T_vexp_p(w_T) angle + langle T_vexp_p(v), T_vexp_p(w_N) angle

::=alphalangle T_vexp_p(v), T_vexp_p(v) angle + langle T_vexp_p(v), T_vexp_p(w_N) angle=langle v, w_T angle + langle T_vexp_p(v), T_vexp_p(w_N) angle.

We must therefore show that the second term is null, because, according to Gauss's Lemma, we must have:

langle T_vexp_p(v), T_vexp_p(w_N) angle = langle v, w_N angle = 0.

* "langle T_vexp_p(v), T_vexp_p(w_N) angle = 0" : Let us define the curve

:alpha : ] -epsilon, epsilon [ imes [0,1] longrightarrow T_pM,qquad (s,t) longmapsto tcdot v(s),with v(0):=v and v'(0):=w_N . We remark in passing that::alpha(0,1) = v(0) = v,qquadfrac{partial alpha}{partial t}(0,t) = v(0) = v,qquadfrac{partial alpha}{partial s}(0,t) = tw_N.

Let us put:

:f : ] -epsilon, epsilon [ imes [0,1] longrightarrow M,qquad (s,t)longmapsto exp_p(tcdot v(s)),

and we calculate:

:T_vexp_p(v)=T_{alpha(0,1)}exp_pleft(frac{partial alpha}{partial t}(0,1) ight)=frac{partial}{partial t}Bigl(exp_pcircalpha(s,t)Bigr)Bigvert_{t=1, s=0}=frac{partial f}{partial t}(0,1)and:T_vexp_p(w_N)=T_{alpha(0,1)}exp_pleft(frac{partial alpha}{partial s}(0,1) ight)=frac{partial}{partial s}Bigl(exp_pcircalpha(s,t)Bigr)Bigvert_{t=1,s=0}=frac{partial f}{partial s}(0,1).Hence:langle T_vexp_p(v), T_vexp_p(w_N) angle = langle frac{partial f}{partial t},frac{partial f}{partial s} angle(0,1).We can now verify that this scalar product is actually independent of the variable t , and therefore that, for example:

:langlefrac{partial f}{partial t},frac{partial f}{partial s} angle(0,1) = langlefrac{partial f}{partial t},frac{partial f}{partial s} angle(0,0) = 0,because, according to what has been given above::lim_{t ightarrow 0}frac{partial f}{partial s}(t,0) = lim_{t ightarrow 0}T_{tv}exp_p(tw_N) = 0being given that the differential is a linear map! This will therefore prove the lemma.
* We verify that "frac{partial}{partial t}langle frac{partial f}{partial t},frac{partial f}{partial s} angle=0" : this is a direct calculation. We first take account of the fact that the maps tmapsto f(s,t) are geodesics, i.e. frac{D}{partial t}frac{partial f}{partial t}=0. Therefore,

:frac{partial}{partial t}langle frac{partial f}{partial t},frac{partial f}{partial s} angle=langleunderbrace{frac{D}{partial t}frac{partial f}{partial t_{=0}, frac{partial f}{partial s} angle+langlefrac{partial f}{partial t},frac{D}{partial t}frac{partial f}{partial s} angle=langlefrac{partial f}{partial t},frac{D}{partial s}frac{partial f}{partial t} angle=frac{partial }{partial s}langle frac{partial f}{partial t}, frac{partial f}{partial t} angle - langlefrac{partial f}{partial t},frac{D}{partial s}frac{partial f}{partial t} angle.Hence, in particular,:0=frac{1}{2}frac{partial }{partial s}langle frac{partial f}{partial t}, frac{partial f}{partial t} angle= langlefrac{partial f}{partial t},frac{D}{partial s}frac{partial f}{partial t} angle=frac{partial}{partial t}langle frac{partial f}{partial t},frac{partial f}{partial s} angle,because, since the maps tmapsto f(s,t) are geodesics, we have langlefrac{partial f}{partial t},frac{partial f}{partial t} angle=mathrm{cste}.

See also

* Riemannian geometry
* Metric tensor


* []

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