# Affine space structure on the space of Hermitian connections

I'm reading Gauduchon's paper Hermitian connections and Dirac operators.

For a fixed almost-Hermitian manifold $$(M, g, J)$$ let $$\mathcal A(g, J)$$ be the space of connections $$\nabla$$ s.t. $$\nabla g = \nabla J=0$$.

The torsion $$T$$ of $$\nabla$$ is seen in the space $$\Omega^2(T M)$$ of vector-valued 2-forms which in turn is seen as a space of trilinear forms that are skew wrt the last two arguments by the identification $$T(X,Y,Z) = g(X, T(Y, Z))$$.

At the beginning of section 2 the paper says $$A(g, J)$$ is an affine space modeled on the space $$\Omega^{1,1}(T M) = \{ B \in \Omega^2(T M): B(JX, JY) = B(X,Y) \}$$.

For a fixed $$\nabla \in \mathcal A(g, J)$$ define its potential wrt the Levi Civita as $$\nabla_X Y - D_X Y =: A^\nabla(X,Y)$$.

Is it that obvious that, given $$\nabla_1, \nabla_2 \in \mathcal A(g, J)$$ then the difference satisfies $$A^{\nabla_1}(JX, JY) - A^{\nabla_2}(JX, JY) = A^{\nabla_1}(X,Y) - A^{\nabla_2}(X,Y) ??$$

Thanks

David

Let $\nabla^0$ be a connection for which $g$ and $J$ are parallel. The way Gauduchon is identifying $\mathcal{A}(g,J)$ with $\nabla^0 + \Omega^{1,1}(TM)$ is by saying that another connection $\nabla$ gives rise to a two form $A^\nabla =\nabla-\nabla^0\in \Omega^2(TM)$, which he views as a trilinear map $X \otimes Y \otimes Z \mapsto g(A^\nabla(X,Y), Z)$. This map is skew symmetric in the last two variables because both $\nabla$ and $\nabla^0$ preserve $g$: $g(\nabla_X Y, Z) = - g(Y, \nabla_X Z) = - g(\nabla_X Z, Y)$.
So we have an element of $\Omega^2(TM)$. It is in these last two variables that we want to check that our form is in $\Omega^{1,1}$. We have $A^\nabla(X,JY) = J A^\nabla (X,Y)$ (since the lack of linearity of $\nabla$ cancels that of $\nabla^0$), and therefore $g (A^\nabla(X,JY), JZ) = g (J A^\nabla(X,Y), JZ) = g (A^\nabla(X,Y), Z)$, where we used the fact that $J$ was hermitian with respect to $g$. It seems to me that you are trying to use $X, Y$ as the skew parameters of the two form instead.
• Why is $A(X, JY) = J A(X, Y)$? To me it seems to be $A(X, JY) = J A(X, Y) - (D_X J)Y$. Anyway it's right that $A^\nabla$ is not skew wrt its two arguments but then maybe the right tensor to take is the torsion. – David P Oct 30 '14 at 14:03
• Also by equations (2.13) and (2.12) notice that using the torsion or $A^\nabla$ is equivalent. – Reimundo Heluani Oct 30 '14 at 14:36
• sorry, I meant $A^\nabla(X, JY) = J A^\nabla(X,Y) - (D_X J)Y$ as $J$ is not $D$-parallel. Anyway, with the definition of $A_{X,Y,Z}$ of formula 2.1.1, what he meant is that if $A'$ is the trilinear form associated to another Hermitian connection $\nabla'$ then the difference $B = A-A'$ satisfies $B_{Z, JX, JY} = B_{Z, X, Y}$. Now it is clear, thanks :) – David P Oct 30 '14 at 14:50
• About taking the torsion, since $T_{X,Y,Z} = (X, T^\nabla_{Y,Z})$, to require $(T-T')_{X,JY, JZ} = (T-T')_{X, Y, Z}$ is the same as saying that $T^\nabla(JY, JZ) - T^{\nabla'}(JY, JZ) = T^\nabla(Y, Z) - T^{\nabla'}(Y,Z)$ – David P Oct 30 '14 at 16:02