# Anti-symmetrization brackets break Einstein summation convention

How does one properly evaluate something of the form $$g_{a}^{\, [b} R_{c] b}~?$$ when I try to expand using the definition of anti-symmetrization brackets the Einstein summation seems to break: $$g_{a}^{\, [b} R_{c] b} = \frac{1}{2} \bigg[ g_{a}^{\, b} R_{c b} - g_a^{\,c} R_{bb}\bigg]$$ the second term should be summing over the $$b$$ index except now I have two lower indices which breaks convention. Is there some implied index raising that's built into the anti-symmetrization brackets? Because I don't see any reason why I should have introduced a metric factor into the second term. It also seems a little fishy since $$b$$ is a dummy index

• Is this from a book? – Qmechanic Mar 22 '19 at 4:14
• @Qmechanic I'm trying to compute the contraction of the Weyl tensor $C_{abc}^{\quad b}$ where $C_{abcd} = R_{abcd} + A(g_{a[d} R_{c] b} + g_{b[c} R_{d] a} ) + B R g_{a [ c} g_{d ] b}$, it's from my course notes – David Feng Mar 22 '19 at 4:16

When contracting $$b$$ with $$d$$, you should have raised the index that was not being covariantly antisymmetrized. For example, instead of the term you showed, write it as $$g_{a[b}{R_{c]}}^b$$.
You cannot (anti) symmetrize indices that belong to different spaces. For example, if you try to symmetrize a mixed tensor like $$A^\mu_{\ \nu}$$ you have to swap $$\mu$$ and $$\nu$$ somehow, but that produces a new tensor with the wrong index placements. The original tensor $$A^\mu_{\ \nu}$$ had $$\mu$$ on top and $$\nu$$ at bottom; trying to symmetrize these generates a new tensor with opposite index placements.