In index notation it makes sense as

$$\sum_j {A_{ij} A_{jk}^T} = \sum_j {A_{ij} A_{kj}}.\tag{1}$$

But this doesn't make sense for Einstein notation where in

$$A^\mu_\sigma (A^\sigma_\nu)^T = A^\mu_\sigma A^\nu_\sigma \tag{2} $$

and the sum is taken over both covariant indices, which is incorrect.

Also how do you know when to include a transpose when going from Einstein notation to matrix notation? For example:

$$\Lambda^\mu_\sigma \eta_{\mu\nu} \Lambda^\nu_\rho = \Lambda^T \eta \Lambda.\tag{3}$$

How do you know that one of the lambdas is transposed?

  • $\begingroup$ So, what does $A\vec{x} = \vec{b}$ look like in Einstein notation? (Hint: $\vec{x}$ is $x^{k}$ in Einstein notation.) What does $\vec{x}^{T} A = \vec{b}^{T}$ look like? (Hint: $\vec{x}^{T}$ is $x_{j}$ in Einstein's notation.) Also note that $A_{\mu\beta}g^{\mu\alpha} = {A^{\alpha}}_{\beta} \neq A_{\beta\mu}g^{\mu\alpha}$ since $A_{\beta\mu}g^{\mu\alpha} = {A_{\beta}}^{\alpha}$, it helps to keep a space to respect the ordering of indices. $\endgroup$ – Alex Nelson Sep 5 '19 at 15:46

Einstein index notation is a form of index notation.

In index notation, the order of upper and lower indices matter, so a notation like $A^\sigma_\nu$ is incorrect. It needs to be either $A^\sigma{}_\nu$ or $A_\nu{}^\sigma$, which are different things. One is the transpose of the other. In your example with the $\Lambda$ matrices, the ambiguity arises because of this incorrect notation.

So if

$$A^\mu{}_\sigma A^\sigma{}_\nu $$

expresses $A^2$, then

$$A^\mu{}_\sigma A_\nu{}^\sigma$$

describes $AA^T$.

  • $\begingroup$ so how you write A inverse in this notation ? $\endgroup$ – Eli Sep 5 '19 at 15:52
  • 1
    $\begingroup$ A inverse is a different matrix, so just something like $(A^{-1})^{\mu}_{\,\,\,\nu}$ $\endgroup$ – Luke Pritchett Sep 5 '19 at 16:34
  • 2
    $\begingroup$ @Eli The inverse of $A^{\mu}_{ \text{ }\text{ }\nu}$ is simply $(A^{-1})^{\sigma}_{ \text{ }\text{ }\rho}$ defined by the relation $(A^{-1})^{\mu}_{ \text{ }\text{ }\sigma}A^{\sigma}_{ \text{ }\text{ }\nu}=\delta^\mu_{\text{ }\text{ }\nu}$. Since the inverse of a matrix is not always trivially expressible in terms of the original matrix, there isn't any index manipulation of the original matrix that can take us to the inverse. $\endgroup$ – Dvij D.C. Sep 5 '19 at 16:35

Can't you write the usual transformation law $$\eta_{ac}{\Lambda^a}_b {\Lambda^c}_d= \eta'_{bd}$$ as
$$ {{(\Lambda^{T})}_b}^a\eta_{ac} {\Lambda^c}_d= \eta'_{bd}$$ if you feel like using matrix notation $\eta'=\Lambda^T \eta \Lambda$?

I think you confuse yourself by writing $\Lambda^a_b$ instead of ${\Lambda^a}_b$.


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