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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 $\nabla =D$?

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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 $d(g(X,Y))(Z)$.

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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

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