I am trying to study the tubular neighborhoods and the normal space/bundle section of Lee's Introduction to Smooth Manifolds.
I have a minor question about the proof of Theorem 6.23 which states: If $M \subseteq \mathbb{R}^n$ is an embedding m-dimensional submanifold, then $NM$ (the tangent bundle) is an embedded n-dimensional submanifold of $T\mathbb{R}^n \approx \mathbb{R}^n \times \mathbb{R}^n$.
Summary: In the proof, Lee first let $E_j|_x = E^i_j(x) \frac{\partial}{\partial x^i}|_x$, where each $E^i_j(x)$ is a partial derivative of $\varphi^{-1}$ evaluated at $\varphi(x)$, $(U, \varphi)$ is a slice chart for $M$ in $\mathbb{R}^n$ centered at $x_0$. $(u^1, \cdots, u^n)$ is the coordinate functions of $\varphi$.
Lee then defined $\varPhi(x, v) = (u^1(x), \cdots, u^n(x), v \cdot E_1|_x, \cdots, v\cdot E_n|_x)$ and calculated its jacobian matrix
$$D\varPhi_{(x,v)} = \begin{pmatrix}
\frac{\partial u^i}{\partial x^j}(x) & 0 \\
* & E^i_j(x)
\end{pmatrix}$$.
My question is: why is the bottom right corner of the Jacobian matrix $E^i_j(x)$? In other words, why is $\frac{\partial v\cdot E_i|_x}{\partial v^j}(x) = E^i_j(x)$?
Here's the screenshot of the theorem and the proof (question highlighted in blue and definitions in yellow):
Best Answer
Well, the matrix of a linear operator is dependent on what basis you choose, so you have to identify the coordinates on the tangent bundle: Say $$v = a^i\frac{\partial}{\partial x^i }$$. Then $(x_1, \ldots, x_n, a^1, \ldots, a^n)$ are coordinates on $T_xU=U \times \mathbb{R}^n$. Then if $$E=E^i_j \frac{\partial}{\partial x^i }$$ then $v \cdot E_j$ (the dot product) in the definition of $\Phi$
$\sum_i a^iE^i_j=a^1E^1_j + a^2E^2_j \cdots$ and so $$\frac{\partial}{\partial a^i }(a^1E^1_j + a^2E^2_j \cdots) = E^i_j$$
Note that we take the derivative with respect to $a^i$ since that coefficient is the coordinate on the tangent space. (I don't fully understand Einstein summation so my indices may be in the wrong position).