Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f:U \to \mathbb{R}^3$ be a surface where $U \subset \mathbb{R}^2$ is open.Let $\Gamma(Tf)$ denote the space of smooth tangent vector fields on $f$

A connection on $f$ is a map $D:\Gamma(Tf) \times \Gamma(Tf) \to \Gamma(Tf)$ with the following properties: $$D_{\alpha X+Y}Z=\alpha D_XY+D_YZ, D_{X}(\beta Y+Z)=\beta D_XY+D_XZ, D_X(kY)=kD_XY+(X^i\frac{\partial k}{\partial u_i})Y$$ for all $\alpha,\beta \in \mathbb{R}$ and smooth maps $k:U \to \mathbb{R}$

The Lie bracket on $\Gamma(Tf)$ (the space of smooth tangent vector fields on $f$ is: $[X,Y]=(X^iY^j_{,i}-Y^iX^j_{,i})f_i$,

here the Eienstein summation convention is used, thus $i,j$ are sumed over $1,2$. Note $X^j_{,i}=\frac{\partial X^j}{\partial u^i}$ and $f_i=\frac{\partial f}{\partial u^i}$,.

Define $T(X,Y):=D_XY-D_YX-[X,Y]$, given that $T$ is a $(1,2)$-tensor field. Prove that if $T=0$ and $D$ preserves the first fundamental form $g$ then $D$ is the same as the covariant differential $\nabla$.

My query:I read from a book (about differential manifolds instead of surfaces) that this means $d(g(X,Y))(Z)=g(D_ZX,Y)+g(X,D_ZY)$, but what is $dg(X,Y)(Z)$? Does one need the uniquess of Levi-Civita connection to prove $\nabla =D$?

share|improve this question

1 Answer 1

up vote 1 down vote accepted
  • Given fields $X$ and $Y$, $g(X,Y)$ denotes a function on your surface, ...

  • ... to which one can apply the operator $d$ of exterior differentiation, which gives as a $1$-form $d(g(X,Y))$ ...

  • which we can apply to vector fields $Z$, to get, again, a scalar function $d(g(X,Y))(Z)$.

share|improve this answer
    
I actually meant that $d(g(X,Y))$ is the exterior differential of the function $g(X,Y)$, just as I wrote. This is a $1$-form, which can be applied to vector fields to give a scalar function. It is a fact, though, that if $f$ is a function and $Z$ a vector field, the value $df(Z)$ of the $1$-form $df$ on the field $Z$ coincides with the result of differentiating $f$ in the direction of $Z$, that is, $(df)(Z)=Zf$. Depending on how you define the exterior differential, this may or may not be part of the definition, in fact. –  Mariano Suárez-Alvarez Sep 24 '12 at 23:36
    
It is in fact the definition of «$g$ is preserved by $D$»! That a connection preserves a metric can also be said that the metric is «parallel with respect to the connection», so that $Dg=0$, for an appropriate definition of how a connection acts on metrics (for comparison, a vector field $X$ is parallel with respect to $D$ if $DX=0$, that is, if $D_YX=0$ for all $Y$); in the end, this equation $Dg=0$ is actuaally the same as the one you quoted. –  Mariano Suárez-Alvarez Sep 25 '12 at 0:16
    
Thanks! Is it possible to use the uniqueness of Levi-Civita connection to prove $\nabla =D$? As the covariant differential $\nabla$ is also compatible with $g$ and torsion free. –  user31899 Sep 25 '12 at 0:24
    
Of course: the Levi-Civita connection is the unique connection whose torsion is zero and for which $g$ is parallel. –  Mariano Suárez-Alvarez Sep 25 '12 at 0:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.