# Einstein Notation question

• I
redtree
TL;DR Summary
I am wondering if I am using it correctly
I am wondering if I am using Einstein notation correctly in the following example.

For a matrix ##R## diagonal in ##1##, except for one entry ##-1##, such as ##R = [1,-1,1]##, is it proper to write the following in Einstein notation:
##R_{\alpha} R_{\beta} = \mathbb{1}_{\alpha \beta} ##, such that ##\Gamma_{\alpha} \Gamma_{\beta} = \Gamma_{\alpha} \Gamma_{\beta} \mathbb{1}_{\alpha \beta} = \big(R_{\alpha}\Gamma_{\alpha} \Big) \Big( R_{\beta}\Gamma_{\beta}\Big)##

LCSphysicist
It is wrong.
First of all, equal index in up and down positions implies summation. So

##R_a R_b = 1_{a b}## Is wrong, and to be honest, i don't even understand what did you was trying to say.

The second equation is also wrong, the index on left hand of the equation is different of the index on right hand side.

• topsquark
redtree
I don't even understand what did you was trying to say.
First off, I am trying to write the following in Einstein notation, where ##R=\mathrm{diag}[1,-1,1]##, then ##R^T R = \mathbb{1}_{\dim{R}}##.

Gold Member
##R## is a matrix and so it has two indices not one. Is the notation ##R = \text{diag} [1,-1,1]## confusing you? They are the entries of the diagonal of the matrix, rather than the components of a vector in some basis. They are the diagonal components of a second rank tensor in some basis. You use ##\delta_{\alpha \beta}## instead of ##\mathbb{1}_{\alpha \beta}##. Anyway, using Einstein's summation convention you would write

$$R^T R= \mathbb{1}$$

as

$$(R^T)^\alpha_{\;\; \gamma} R^\gamma_{\;\; \beta} = \delta^\alpha _{\;\; \beta} .$$

or

$$R^{\;\; \alpha}_ \gamma R^\gamma_{\;\; \beta} = \delta^\alpha _{\;\; \beta} .$$

or as ##R## is symmetric

$$R^\alpha_{\;\; \gamma} R^\gamma_{\;\; \beta} = \delta^\alpha _{\;\; \beta} .$$

I suppose, it's not standard, you could introduce the numbers ##r_{(1)}= 1 , r_{(2)} = -1, r_{(3)} = 1## where I have used brackets around the indices to indicate that I'm simply taking them to be numbers rather than components of a tensor in some basis, which is what they actually are. And then write

$$R^\alpha_{\;\; \beta} = r_{(\beta)} \delta^\alpha_{\;\; \beta}$$

no summation over ##\beta## is implied. Then you could write

\begin{align*}
R^\alpha_{\;\; \gamma} R^\gamma_{\;\; \beta} & = r_{(\alpha)} r_{(\beta)} \delta^\alpha_{\;\; \gamma} \delta^\gamma_{\;\; \beta}
\nonumber \\
& = r_{(\alpha)} r_{(\beta)} \delta^\alpha_{\;\; \beta}
\nonumber \\
& = r_{(\alpha)}^2 \delta^\alpha_{\;\; \beta}
\nonumber \\
& = \delta^\alpha_{\;\; \beta}
\end{align*}

But this is an abuse of the usual convention.

Last edited:
• vanhees71, topsquark and redtree