4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 '14 at 14:03
  • $\begingroup$ Also by equations (2.13) and (2.12) notice that using the torsion or $A^\nabla$ is equivalent. $\endgroup$ – Reimundo Heluani Oct 30 '14 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 '14 at 14:50
  • $\begingroup$ You're right, I forgot to take the difference of the two forms :) $\endgroup$ – Reimundo Heluani Oct 30 '14 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 '14 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.