Help in proof from Riemannian Geometry by Docarmo.

Question

I have been working on ${\it Lemma\,5.2}$ from Riemannian Geometry by DoCarmo which establishes the existence and uniqueness of the vector field $Zf=(XY-YX)f$, given $X$ and $Y$ as differenciable vector fields. On this proof we have expressions for $XYf$ and $YXf$ as follows:

• $XYf=\sum_{i,j}a_{i}\frac{\partial b_{j}}{\partial x_{i}}\frac{\partial f}{\partial x_{j}}+\sum_{i,j}a_{i}b_{j}\frac{\partial^{2}f}{\partial x_{i}\partial x_{j}}$
• $YXf=\sum_{i,j}b_{j}\frac{\partial a_{i}}{\partial x_{i}}\frac{\partial f}{\partial x_{j}}+\sum_{i,j}a_{i}b_{j}\frac{\partial^{2}f}{\partial x_{j}\partial x_{i}}$

Where $Xf=\sum_{i}a_{i}\frac{\partial f}{\partial x_{i}}$ and $Yf=\sum_{j}b_{j}\frac{\partial f}{\partial x_{j}}$. If I substract expressions of the items I obtain $$Zf=XYf-YXf=\sum_{i,j}\left(a_{i}\frac{\partial b_{j}}{\partial x_{i}}\frac{\partial f}{\partial x_{j}}-b_{j}\frac{\partial a_{i}}{\partial x_{j}}\frac{\partial f}{\partial x_{i}}\right).$$But DoCarmos says that this turns out to be $$Zf=XYf-YXf=\sum_{i,j}\left(a_{i}\frac{\partial b_{j}}{\partial x_{i}}-b_{i}\frac{\partial a_{j}}{\partial x_{i}}\right)\frac{\partial f}{\partial x_{j}}$$ as if $\frac{\partial f}{\partial x_{i}}$ and $\frac{\partial f}{\partial x_{j}}$ were the same.

Answer 1

They are nearly the same, up to indices. Try do like this: $$ Zf=XYf-YXf=\sum_{i,j}\left(a_{i}\frac{\partial b_{j}}{\partial x_{i}}\frac{\partial f}{\partial x_{j}}-b_{j}\frac{\partial a_{i}}{\partial x_{j}}\frac{\partial f}{\partial x_{i}}\right) = \sum _{i,j} a_i\partial_i b_j \partial_j f - \sum_{i,j} b_j \partial_j a_i \partial_i f \stackrel ! = \sum_{i,j} a_i \partial_i b_j \partial_j f - \sum_{i,j} b_{\color{red}{i}} \partial _{\color{red}{i}} a_{\color{blue}{j}}\partial_{\color{blue}{j}} f = \sum_{i,j} (a_i \partial_i b_j - b_i \partial_i a_j)\partial_j f. $$ or $$ Zf = \sum_{i,j} (a_j \partial_j b_i - b_j \partial_j a_i)\partial_i f. $$

The answer is $$ \newcommand \i{\color{red}i} Z f = \sum_{i,j} \left(a_j\frac {\partial b_{\i}}{\partial x_j} - b_j \frac {\partial a_{\i}}{\partial x_j}\right) \frac{\partial f}{\partial x_{\i}}. $$