How do you write $A A^T$ in Einstein notation?

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?

• 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. – 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 $$AA^T$$.

• so how you write A inverse in this notation ? – Eli Sep 5 '19 at 15:52
• A inverse is a different matrix, so just something like $(A^{-1})^{\mu}_{\,\,\,\nu}$ – Luke Pritchett Sep 5 '19 at 16:34
• @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. – Dvij Mankad 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 $${\Lambda^a}_b$$.