Why does g_{αβ}ϵ^{αβγδ} equal zero?

[bludragn0]

Homework Statement

Hi all, I'm having trouble evaluating the product $g_{αβ}ϵ^{αβγδ}$. Where the first term is the metric tensor and the second is the Levi-Civita pseudotensor. I know that it evaluates to 0, but I'm not sure how to arrive at that.

The Attempt at a Solution

My first thought process was that every permutation will include at least 2 indices which are equal, (because every permutation will have at least 2 zeroes) which makes every term zero. That seems too trivial however. Sorry if this sounds totally nonsensical, but I haven't been able to find a resource that really clarifies this.

 

[tman12321]

I think you are close to the right answer, but I can't tell because your language is a little too vague. Permutations in which indices? The summation occurs only over the α and β, so these are the only indices you need to worry about. Recall as well that g_{αβ} is symmetric, and if ε^{αβγδ} ≠ 0, then ε^{αβγδ} = -ε^{βαγδ}. Also, ε^{αβγδ} = 0 if any of the α,β,γ,δ are equal.

 

[dauto]

That's very simple. The metric is symmetric and the Levi-Civita is anti-symmetric.
g_αβ = g_βα, and
ε^αβγδ = -ε^βαγδ
g_αβε^αβγδ = g_βα(-ε^βαγδ) = - g_αβε^αβγδ