1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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 $AA^T$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 ${\Lambda^a}_b$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.