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

Question

In index notation it makes sense as

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

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

$$A^\mu_\sigma (A^\sigma_\nu)^T = A^\mu_\sigma A^\nu_\sigma $$

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:

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

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

Answer 1

Follow this hint: $$(AA^{\top})^i{}_j=A^i{}_s(A^{\top})^s{}_j$$ $$\qquad =A^i{}_s\ A^j{}_s$$

You can verify for with $$ \left[\begin{array}{cc} A^1{}_1&A^1{}_2\\ A^2{}_1&A^2{}_2 \end{array}\right] \left[\begin{array}{cc} A^1{}_1&A^2{}_1\\ A^1{}_2&A^2{}_2 \end{array}\right]= \left[\begin{array}{cc} A^1{}_1A^1{}_1+A^1{}_2A^1{}_2&A^1{}_1A^2{}_1+A^1{}_2A^2{}_2\\ A^2{}_1A^1{}_1+A^2{}_2A^1{}_2&A^2{}_1A^2{}_1+A^2{}_2A^2{}_2 \end{array}\right]$$

Answer 2

To define $A^{\mathsf{T}}$ within the Ricci calculus, we need a nondegenerate symmetric bilinear form $g$: $$g(x,Ay)=g(A^{\mathsf{T}}x,y)\text{.}$$ Note that the definition of $(\quad)^{\mathsf{T}}$ depends on the $g$ we choose. If the components of $g$ are $g_{ij}$ and those of its dual are $g^{ij}$, then $${{(A^{\mathsf{T}} )}^i}_j=g^{il}{A^{k}}_lg_{kj}$$ $${{(AA^{\mathsf{T}})}^{i}}_j={A^i}_m{{(A^{\mathsf{T}} )}^m}_j=g_{ik}g^{jl}{A^i}_j{A^k}_l\text{.}$$

"$\Lambda^{\mathsf{T}}\eta\Lambda$" is an abus de notation that I wish physicists would avoid.