I am trying to understand the blow-up along a submanifold as explained in Huybrechts "Complex Geometry An Introduction", p. 99, Example 2.5.2.
For background, for $m \leq n$, we see $\mathbb C^m \subset \mathbb C^n$ as $\{ z_{m+1} = \ldots = z_n = 0 \}$ and define $$ Bl_{\mathbb C^m} (\mathbb C^n) := \{ (x,z): z_i x_j = z_j x_i, \forall i, j = m+1, \ldots, n \} \subset \mathbb P^{n-m-1} \times \mathbb C^n $$
Let $U, V \mathbb C^n$ be two open sets and with a biholomorphic map $\phi: U \rightarrow V$ with $\phi(U \cap \mathbb C^m) = V \cap \mathbb C^m$. I am trying to understand Huybrechts' explanation of how to glue together $Bl_{\mathbb C^m \cap U} (U)$ and $Bl_{\mathbb C^m \cap V}(V)$ using $\phi$.
Firstly, writing the coordinate in $U$ as $z=(z_1, \ldots, z_n)$ and $\phi=(\phi^1, \ldots, \phi^n)$, he defines the complex numbers $\phi_{k,j}$ by the equation $$ \phi^k = \sum_{j=m+1}^n z_j \phi_{k,j} $$ This seems a bit nonsensical to me, as $\phi^k$ is simply a function of $z$, it's not necessarily linear in $z$. But what I think he means in fact is this: if we write $z_i' := \phi^i(z)$, then: $$ \frac{\partial}{\partial z_k'} = \sum_{j=m+1}^n \frac{\partial \phi^k}{\partial z_j} \frac{\partial}{\partial z_j} $$ so then $\phi_{k,j} = \frac{\partial \phi^k}{\partial z_j}$ in fact. Now he says that the map $\hat \phi: \mathbb P^{n-m-1} \times (\mathbb C^n \cap U) \rightarrow \mathbb P^{n-m-1} \times (\mathbb C^n \cap V)$ given by: $$ \hat\phi(x,z) := ((\phi_{k,j}(z))_{k,j=m+1,\ldots, n} \cdot x, \phi(z)) $$ defines a map $\hat\phi: Bl_{\mathbb C^m \cap U} (U) \rightarrow Bl_{\mathbb C^m \cap V} (V)$ when restricted to $Bl_{\mathbb C^m \cap U} (U)$.
I don't understand why $\hat\phi(Bl_{\mathbb C^m \cap U} (U)) \subset Bl_{\mathbb C^m \cap V} (V)$.
Indeed, the inclusion would mean that if $z_i x_k = z_k x_i$ for $i,k = m+1, \ldots, n$, then: $$ \phi^i(z) \sum_{j={m+1}}^n \frac{\partial \phi^k}{\partial z_j}(z) x_j = \phi^k(z) \sum_{j={m+1}}^n \frac{\partial \phi^i}{\partial z_j}(z) x_j $$
As it stands, this condition seems blatantly false!
So what am I getting wrong here? Maybe this is not the condition that needs to be checked? But then how are $Bl_{\mathbb C^m \cap U} (U)$ and $Bl_{\mathbb C^m \cap V}(V)$ glued using $\phi$?
First of all, in the equations for $k\geq m+1$, $$ \phi^k(z) = \sum_{j=m+1}^n z_j \phi_{k,j}(z), \tag{1} $$ the author didn't mean that $\phi_{k, j}(z)$ are numbers. (His omitting $(z)$ seems to be the source of the confusion.) They are actually functions of $(z_1,\dots,z_n)$, and their existence is guaranteed by the condition that $\phi(U\cap {\mathbb C}^m) = V\cap {\mathbb C}^m$. Basically, when $z_j=0$ for all $j\geq m+1$ we have $\phi^k=0$ for all $k\geq m+1$, so we have such a form. (See also the proof of him 2.4.7.)
The $\phi_{k,j}$ are different from your linearized version $\frac{\partial \phi^k}{\partial z_j}$, although they are the same on $U\cap {\mathbb C}^m$, where $z_{m+1}=\dots=z_n=0$. For we can differentiate (1) to see \begin{align*} \frac{\partial \phi^k}{\partial z_j} &= \frac{\partial}{\partial z_j}\Big(\sum_{l=m+1}^n \phi_{k,l}(z) z_l\Big)\\ &= \phi_{k, j}(z) + \sum_{l=m+1}^n \frac{\partial \phi_{k,l}}{\partial z_j} z_l. \end{align*} Therefore, on $U\cap {\mathbb C}^m$ when $z_{m+1}=\dots=z_n=0$, we have $$ \frac{\partial \phi^k}{\partial z_j}(z) =\phi_{k, j}(z). $$
Then the condition you proposed to check is totally right. That is, if $$ z_ix_k = z_kx_i\tag{2} $$ for $i, k=m+1,\dots,n$, we should have $$ \phi^i(z)\sum_{j=m+1}^n \phi_{k,j}(z)x_j = \phi^k(z)\sum_{j=m+1}^n \phi_{i,j}(z)x_j. $$
The proof is simply plugging in the equation (1) and use the incidence relation (2). \begin{align*} \phi^i(z)\sum_{j=m+1}^n \phi_{k,j}(z)x_j &\overset{(1)}{=} \sum_{l=m+1}^n \phi_{i, l}(z)z_l \sum_{j=m+1}^n \phi_{k,j}(z)x_j\\ &=\sum_{j,l=m+1}^n \phi_{i, l}(z) \phi_{k,j}(z) z_l x_j\\ &\overset{(2)}=\sum_{j,l=m+1}^n \phi_{i, l}(z) \phi_{k,j}(z) z_j x_l\\ &=\sum_{j=m+1}^n \phi_{k,j}(z)z_j \sum_{l=m+1}^n \phi_{i, l}(z)x_l\\ &\overset{(1)}{=}\phi^k(z) \sum_{j=m+1}^n \phi_{i,j}(z)x_j. \end{align*}