0

Let $g^{ij}$ be a symmetric $(2,0)$ tensor and $F_{ij}=F_{[ij]}$, then what can we say about the product $g^{ij}F_{ij}$? Does it always have the same sign, is it always zero or it depends on the specific $g$ and $F$? (About $F_{[ij]}$ : Any rank $2$ tensor can be written as $F_{ij}=\frac{F_{(ij)}+F_{[ij]}}{2}$, where $F_{(ij)}$ is a symmetric tensor and $F_{[ij]}$ is an antisymmetric tensor.)

JBuck
  • 681

1 Answers1

1

While $g^{ji}F_{ji}=g^{ij}F_{ij}$ by relabelling, on your assumptions $g^{ji}F_{ji}=-g^{ij}F_{ij}$ by exchanging indices. Combining these, $2g^{ij}F_{ij}=0$. The only way to avoid $g^{ij}F_{ij}=0$ is to work in characteristic $2$.

J.G.
  • 115,835