I have a transition map $\phi _i \circ \phi ^{-1}_j (u) = (\frac{u^1}{u^{i+1}},\frac{u^2}{u^{i+1}},...,\frac{u^i}{u^{i+1}},\frac{u^{i+2}}{u^{i+1}},...,\frac{u^j}{u^{i+1}},\frac{1}{u^{i+1}},\frac{u^{j+1}}{u^{i+1}},...,\frac{u^n}{u^{i+1}})$. I want to find the Jacobian matrix. And more importantly, the determinant of the matrix so I can determine orientability.
I understand I am supposed to differentiate the first coordinate partially wrt $u^1,...,u^n$ to give the first row of the matrix, and do the same to the second coordinate to give the second row of the matrix etc. However, where I am a bit confused is how to treat the $i$'s of each coordinate. Should I just give the i a value so I can carry out the differentiation?
Your help would be greatly appreciated.
I doubt anyone has ever done this calculation explicitly. Of course, what you are talking about is the transition functions for the natural coordinate charts on the projective space $P^n$. This transition function is between the $j$-th chart and the $i$-th chart. Here $U_j=\{[x^0,\dots,x^n]:x^j\neq 0\}\subset P^n$, and $$ \phi_j: U_j\to {\mathbb R}^n;\ [x^0,\dots,x^n]\mapsto \big(\frac{x^0}{x^j},\dots,\hat{\frac{x^j}{x^j}},\dots,\frac{x^n}{x^j}\big), $$ where the hat means omitting.
Even working out the transition function is confusing, but your formula was perfect. So for concreteness, we are talking about the transition $\phi_i\circ \phi_j^{-1}$ for $0\leq i<j\leq n$. ($i=j$ would give the identity, and $i>j$ can be similarly treated.)
Let me call $\phi_i\circ \phi_j^{-1}$ by $f$ and its components $f^k$, so $$ f(u) =\phi_i\circ \phi_j^{-1}(u)=(f^1(u),\dots, f^n(u)). $$ The Jacobian matrix is then $$ J(f)=\big(\frac{\partial f^k}{\partial u^\ell}\big)_{n\times n}. $$ One important point is that $$ u^{i+1} f(u) = g(u) = (u^1,\dots,u^i, u^{i+2},\dots,u^j,1,u^{j+1},\dots,u^n). $$ The Jacobi matrix of $g$ is then $$ J(g)=E_{1,1}+\dots+E_{i,i}+E_{i+1,i+2}+\dots+E_{j-1,j}+E_{j+1,j+1}+\dots+E_{n,n}, $$ where $E_{a,b}$ is the matrix with only one nonzero entry 1 at position $(a, b)$. So $J(g)$ is obtained from the identity matrix by making its $j$-row zero, and shifting the 1's in the $(i+1)$ to $(j-1)$-st rows to the right by one position. Therefore its $i+1$st column is also zero. (These just say that since $g_j=1$, the $j$-th row of $J(g)$ is zero, and since $g$ does not involve $u^{i+1}$, the $(i+1)$-st column of $J(g)$ is zero.)
By the product rule, we have from $f=\frac{1}{u^{i+1}}g$ that $$ J(f)= \frac{1}{u^{i+1}}J(g) + g^T\cdot D\frac{1}{u^{i+1}}, $$ where $D\frac{1}{u^{i+1}}=(0,\dots,0,-1/(u^{i+1})^2,0,\dots,0)$ is the Jacobian matrix of $\frac{1}{u^{i+1}}$, and $g^T$ is its transpose as a column vector. So $g^T\cdot D\frac{1}{u^{i+1}}$ has only one nonzero column, the $(i+1)$-st.
Since $J(g)$ has exactly a one in each row and column except in the $(j, i+1)$ position, there is only one nontrivial term in the determinant of $J(f)$. That is, we need to use the $(j,i+1)$ entry of $g^T\cdot Du^{i+1}$, which is $1\cdot(-1/(u^{i+1})^2)$, noting $g^{j}=1$.
Putting these together, and noting the $\frac{1}{u^{i+1}}$ in front of $J(g)$ in $J(f)$, we see that $$ \det J(f) = \pm \frac{1}{(u^{i+1})^{n+1}}, $$ and we just need to determine the sign now.
By the above analysis, the sign is related to the permutation matrix of the cycle $(i+1, i+2, \dots, j)$, and its sign is $(-1)^{j-i-1}$. Noting that the entry $J(f)_{j,i+1}=1\cdot(-1/(u^{i+1})^2)$ has one more $-1$, so the final answer is $$ \det J(\phi_i\circ\phi_j^{-1}) = \frac{(-1)^{j-i}}{(u^{i+1})^{n+1}}, \quad 0\leq i<j\leq n. $$
Of course, I have the interest to work this out only after my computer calculation shows it. To be complete, we also record that when $i>j$, we have $$ \det J(\phi_i\circ\phi_j^{-1}) = \frac{(-1)^{j-i}}{(u^{i})^{n+1}}, \quad 0\leq j<i\leq n, $$ and $\phi_i\circ \phi_j^{-1}=Id$ if $i=j$.