# What is the value of the 2-form $X(u(Y))-Y(u(X))-u([X,Y])$, is it $2du(X,Y)$?

Let $X,Y$ be two vector fields on a Riemannian manifold $(M,g,\nabla)$ where $\nabla$ is the symmetric metric connection on $M$ and let $u$ be a 1-form. I want to find the value or a simple form or interpretation of the following $$(1)\hspace{.51cm} X(u(Y))-Y(u(X))-u([X,Y])\\ (2)\hspace{.51cm} X(u(Y))-u(\nabla_XY)-u(X)u(Y)$$ John show that the first expression is a 2-form, I have been trying to prove that it is $2du$ as he claimed. We can take $X=\Sigma X^i \frac{\partial}{\partial x_i}$ , $Y=\Sigma Y^i \frac{\partial}{\partial x_i}$ and $u(\frac{\partial}{\partial x_i})=u_i$

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Expression (1) is not in general identically equal to zero. Indeed we have, for any $1$-form $u$,

$du(X, Y) = X(u(Y)) - Y(u(X)) - u([X, Y]), \tag{1}$

where $X$ and $Y$ are vector fields on $M$. (1) is a well-known result and is the $p = 1$ case of a broader formula expressing $d\omega$ for any $p$-form $\omega$. For more information about (1) and related matters see http://en.m.wikipedia.org/wiki/Exterior_derivative.

As for the second expression,

$X(u(Y)) - u(\nabla_X Y) - u(X) u(Y), \tag{2}$

the best I can do is note that

$X(u(Y)) = \nabla_X (u(Y)) = (\nabla_X u)(Y) + u(\nabla_X Y), \tag{3}$

which holds since $u(Y)$ is a function and the differential operators $X$ and $\nabla_X$ agree on functions. In light of (3), (2) becomes

$X(u(Y)) - u(\nabla_X Y) - u(X) u(Y) = (\nabla_X u)(Y) + u(\nabla_X Y) - u(\nabla_X Y) - u(X) u(Y) = (\nabla_X u)(Y) - u(X) u(Y); \tag{4}$

I don't think there is any obvious, simple way to simplify (4) further since $u(X) u(Y)$ is "quadratic", that is, of second "degree" in $u$; since $\nabla_X$ is a linear operator, most of the standard identities don't address products such as $u(X) u(Y)$. So that's about as fat as I can take things.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Uh, Finally I found a reference for the first identity, Bishop and Goldberg, tensor analysis on manifolds, p 169. Unfortunately, he put a factor on the RHS and gave a reason that I did not understand again but I accept it as it is now. Thanks for great help. –  Sameh Shenawy Jan 26 '14 at 21:23
The reason is: The annoying factor can be eliminated by using another definition of wedge products. This alternative definition, which does not alter the essential properties of wedge product, is obtained by magnifying our present wedge product of a p-form and a q-form by the factor $(p + q)!/p!q!$. Both products are in common use and we shall continue with our original definition.) –  Sameh Shenawy Jan 26 '14 at 21:30
@ Semsem: the identity in question occurs in a great many books on differential geometry and topology etc., but I didn't have any of mine on hand when I wrote the answer, so I resorted to googling up what I could. Glad to help out. Thanks for the kind words, the vote of confidence, and of course the "acceptance"! –  Robert Lewis Jan 26 '14 at 21:37

For the first one, note that if we consider $X\mapsto fX$ for some function $f$, then

$$(fX)\big(u(Y)\big) - Y\big(u(fX)\big) - u([fX, Y]) = f\big(X\big(u(Y)\big) - Y\big(u(X)\big) - u([X, Y])\big).$$

Together with the fact that the expression is antisymmetric in $X$ and $Y$, we see that the first expression is a two form. Putting $X = \frac{\partial}{\partial x_i}$ and $Y=\frac{\partial}{\partial x_j}$ shows that the first expression is $2du$.

For the second one, the first two terms can be written $\nabla u(X, Y)$. Not sure what to do with the last term.

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Thanks. Do you mean by $(\nabla^2u)(X,Y)$ the hessian $Hess_u(X,Y)$of $u$ –  Sameh Shenawy Jan 26 '14 at 10:04
@Semsem: Yes. That's the hessian. –  John Ma Jan 26 '14 at 10:05
$Hess_f(X,Y)=X(Y(f))-(\nabla_XY)f$ which $f$ you are consider –  Sameh Shenawy Jan 26 '14 at 10:11
the first term is $X(u(Y))$ not $X(Y(u))?$ on the other hand the last one is not defined as I understand, thank you again @John –  Sameh Shenawy Jan 26 '14 at 10:14
@Semsem: My bad -- I thought $u$ was a function ^^. Let me withdraw the answer. –  John Ma Jan 26 '14 at 10:58