How do you prove this tensor identity? Suppose we have a tensor field ${\Gamma_{ab}}^{c}$ on a smooth manifold $M$ such that for all vector fields $X^a$, we have $X^aX^b{\Gamma_{ab}}^{[c}X^{d]} = 0$.
How do you show that there exists a 1-form field $\Upsilon_a$ such that ${\Gamma_{(ab)}}^c = \Upsilon_a {\delta_b}^c + \Upsilon_b {\delta_a}^c$, where ${\delta_a}^b$ is the identity endomorphism of $TM$?  
 A: The first observation is that the assumption and the conclusion does not depends on differentiation, thus this is a pointwise statement and has nothing to do with the smooth manifold $M$. We just need to prove the statement at a point. 
Now we think of $\Gamma$ as a multilinear map 
$$ \Gamma : V\times V \to V.$$
Then the condition is the same as saying that 
$$\Gamma(X, X) \otimes X = X\otimes \Gamma(X, X).$$
Thus there is $\lambda :V\to \mathbb R$ so that 
$$\Gamma(X, X) = \lambda (X) X$$
for all $X\in V$. Note that one has $\lambda(aX) = a\lambda (X)$. 
Now we investigate the property of $\lambda$. Now write $S(X, Y) = \Gamma(X, Y) + \Gamma(Y, X)$, then we have 
$$(*)\ \ S(X, Y)= (\lambda(X+Y) - \lambda(X))X + (\lambda(X+Y) - \lambda(Y))Y.$$
If $\lambda$ is linear, then the require expression is shown by letting $\Upsilon = \lambda$. 
Claim $\lambda$ is linear. 
From $(*)$, using $X\mapsto -X$, we have 
$$ -S(X, Y)= (\lambda(-X+Y) - \lambda(-X))(-X) + (\lambda(-X+Y) - \lambda(Y))Y.$$
Comparing coefficient of $Y$ gives 
$$\lambda(X+Y) - \lambda(Y) = -\lambda(-X +Y) + \lambda(Y), $$
which is the same as 
$$\lambda(X+Y) + \lambda(Y-X) = 2\lambda(Y),$$
Put $X = \frac{v-u}{2}, Y = \frac{v+u}{2}$, then 
$$\lambda (v) + \lambda(u) = \lambda(u+v).$$
Thus $\lambda$ is linear. 
