Some questions about Hitchin's self-duality paper I am reading this paper (The self-duality equations on a Riemann surface by N. Hitchin), and I don't understand a few things in page 67. In proof of Theorem 2.1 after Equation 2.4, he gives the relation for $F(B)s$ as
$$F(B)s=F(A)s-(\deg L)\omega s+ \frac12 \deg(∧^2V)\omega s.$$ 
In the proof, he obtains a connection $B$ on $L^*V$ using a connection $A$ on $V$, and a connection with curvature $(\deg L)\omega$. I am thinking of $L^*V$ as a pull back of $V$, or am I wrong? If I understand what is the explicit form of this connection, I may get the curvature $F(B)$, and eventually above relation. Or is there any other way to get the above result? 
I tried a lot but couldn't understand how he obtains the result above for $F(B)s$. This formula is very essential in the proof, because he uses this information in Theorem 2.1 to give a condition for stability that is discussed in the third section of the paper. 
I will be very glad if somebody could explain me how to derive this result. 
I have another small question which maybe related to the above question: 
Initially in the top of page 67 he says "Fixing a connection $A_0$ on $∧^2V$, we find that a connection $A$ on $P$ lifts to $\tilde{P}$  whose curvature is $F(A)+1/2F(A_0)1$". Here I don't see how he can get this curvature for the lift. I was thinking, maybe he uses this curvature to find the above formula. Can someone elaborates on this one as well? 
I will be very grateful for your help. 
 A: Using your connection for the tensor product, we can find the curvature as
$$F_{\nabla\otimes\bar{\nabla}}(s_1\otimes s_2) =\nabla\otimes\bar{\nabla}\big(\nabla\otimes\bar{\nabla}(s_1\otimes s_2)\big)=\nabla\otimes\bar{\nabla}\big(\nabla(s_1)\otimes s_2
+s_1\otimes \bar{\nabla}(s_2)\big)=\nabla^{2}(s_1)\otimes s_2
-\nabla(s_1)\otimes \bar{\nabla} (s_2)+\nabla(s_1)\otimes \bar{\nabla} (s_2)+s_1\otimes \bar{\nabla}^{2}(s_2)\\= \nabla^{2}(s_1)\otimes s_2+s_1\otimes \bar{\nabla}^{2}(s_2)$$
the minus signs appears because $\nabla(s_1)\otimes s_2$ is (1,0) form. 
Next we notice that $A_0$ is a fixed connection on $\wedge^2 V$ and on $V$ we have the connection $A+A_0$.Thus, the covariant derivative for connection $A_0+A$ is given by $d_{A+A_0}=d+A+A_0$, so the curvature on $V$ would be 
$$F_{A+A_0}= d^2_{A+A_0}=dA+A\wedge A+A_0\wedge A_0\\=dA+A\wedge A+\frac{1}{2}[A_0,A_0]$$
Defining $F(A):=dA+A\wedge A$ and $F(A_0):=[A_0,A_0]$, we get the curvature on $V$ according to $F(A)+\frac{1}{2}F(A_0)$ (This is the equation given in the first paragraph of page 67) Is this derivation of $F(A)+\frac{1}{2}F(A_0)$ correct? 
Furthermore, since $\wedge^2 V$ is a line bundle embedded in $V$, its curvature has the form $F(A_0)=deg(\wedge^2 V)\omega$ where $\omega$ is a positive form. 
So now we have 
$$F_{B}(s_1\otimes s_2)=F(V)(s_1)\otimes s_2+s_1\otimes F(L^*)(s_2)\\=(F(A)+\frac{1}{2}deg(\wedge^2 V)\omega)s_1\otimes s_2+s_1\otimes(-(deg L)\omega) s_2 \\=F(A)(s_1\otimes s_2)+\frac{1}{2}deg(\wedge^2 V)\omega (s_1\otimes s_2) - (deg L)\omega(s_1\otimes s_2)$$
Above, I used the fact that for a line bundle $L$ with some connection, the curvature is given by $(deg L)\omega$ where $\omega$ is a positive form. The minus sign is because we look at the dual space, $L^*$.
Here, I assume that in paper the section $s\in \Omega^0(M, L^*\otimes V)$ must be of the form $s:=s_1\otimes s_2$. Am I right?
If you see any errors/problems with my computations, please let me know. I will be very thankful for that!
