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 $$ A^{\nabla_1}(JX, JY) - A^{\nabla_2}(JX, JY) = A^{\nabla_1}(X,Y) - A^{\nabla_2}(X,Y) ?? $$




1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – David P
    Oct 30, 2014 at 14:03
  • $\begingroup$ Also by equations (2.13) and (2.12) notice that using the torsion or $A^\nabla$ is equivalent. $\endgroup$ Oct 30, 2014 at 14:36
  • $\begingroup$ 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 :) $\endgroup$
    – David P
    Oct 30, 2014 at 14:50
  • $\begingroup$ You're right, I forgot to take the difference of the two forms :) $\endgroup$ Oct 30, 2014 at 14:55
  • $\begingroup$ 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 $T^\nabla(JY, JZ) - T^{\nabla'}(JY, JZ) = T^\nabla(Y, Z) - T^{\nabla'}(Y,Z)$ $\endgroup$
    – David P
    Oct 30, 2014 at 16:02

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