# General form of a metric affine connection with zero curvature

I have read somewhere that the general form of a metric affine connection whose Riemann curvature is zero is given by $$\Gamma^{i}_{ij}=A^{i}_{\alpha}\partial_{j}A^{\alpha}_{k}$$ where $A^{i}_{\alpha}$ are elements of some invertible matrix. This form comes from a general solution of the partial differential equations

$$R^{i}_{jkl}\equiv 2(\partial_{j}\Gamma^{i}_{kl}-\Gamma^{i}_{kp}\Gamma^{p}_{jl})_{[jk]}=0.$$ How does the above form of the connection coefficients is a general solution to the system of pde's? Substitution of the expression is obviously a proof, but where does that form come from?

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Actually, your formula is not correct because you fail to distinguish between $A$ and $A^{-1}$, probably because you are not familiar with the upper and lower index conventions. Also, you have a typo where you have '$ij$' in the left hand side subscript when you mean '$jk$'. If you set $B = A^{-1}$, then the formula should read $$\Gamma^i_{jk} = B^i_{l}\partial_jA^l_k\ .$$ Your question amounts to asking why the vanishing of the curvature of a connection $\nabla$ is equivalent to the (local) existence of a basis of $\nabla$-parallel vector fields. This is not a research question; read a book. –  Robert Bryant Sep 9 '13 at 9:16
Addendum: I have to apologize if my last sentence in the above comment sounded rather dismissive; I ran out of space and time in writing the comment and could only now get back to it. If I could, I would replace the last sentence with something like, "This is not a research question, as this is treated in nearly every introductory discussion of curvature and connections. You'll get much better information and understanding by reading the treatment in any standard book on differential geometry (cf. Volume II of Spivak) than you will from a few paragraphs that someone will write on MO." –  Robert Bryant Sep 9 '13 at 11:49
Sorry Sir for the typos. I'll consult a book. Thank you for your response. –  AyanVivek Sep 9 '13 at 15:34