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Suppose we have a smooth manifold $M$, and one point $p\in M$. Let $U$ be an open set around $p$ that admits two parametrizations, namely $x:V_1\to U$ and $y:V_2\to U$. Now in these two parametrizations we can write vector field $X$ on $M$ as

$X=\sum_{i=1}^n{a_i(p)\frac{\partial}{\partial x_i}}=\sum_{i=1}^n{b_i(p)\frac{\partial}{\partial y_i}}.$

Suppose now that we have a smooth function $f:M\to R$. Then value $Xf$ should not depend on the parametrization. So, I am trying to prove that.

Denote by $A(p)=(a_1(p),...,a_n(p))$, similarly for $B(p)$ and let $M=D(y^{-1}\circ x)(x^{-1}(p))$. Then we have $B(p)=M\cdot A(p)$. Denote by $\nabla_Xf$ and $\nabla_Yf$ gradients of $f$ with respect to two parametrizations. Then we get:

$Xf=A(p)\cdot \nabla_Xf$ for the first parametrization and

$Xf=B(p)\cdot \nabla_Yf=M\cdot A(p) \cdot \nabla_Xf\cdot M^{-1}$, for the second parametrization.

However these two are not equal. Could you please point out the mistake?

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    So, $M=D(y^{-1}\circ x)(x^{-1}(p))$, is this the problem?2012-09-27

1 Answers 1

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You're not consistent in your use of row and column matrices. Your first expression for $Xf$ is a $1 \times 1$ matrix, but the second one is not even defined! (The matrix product $M \cdot A(p)$ is undefined if $A(p)$ is a row vector. Similarly for $\nabla_X f \cdot M^{-1}$.)

Try adding a few transposes in the right places; then it will work better, because of the rule $(AB)^t = B^t A^t$.