# Einstein notation: can a free index be upper in one term and lower in another term?

Consider a linear combination of terms written using Einstein notation.

Consider one free index in the linear combination: is it necessary that the index is upper in all terms or lower in all terms, or not?

Equivalently is it possible to have the same free index, but in one term it is a upper index and in the other it is a lower index?

For example, is it possible to have the following expression?

$$A_\mu+B^\mu$$

Or this is forbidden in einstein notation, and one free index can only appear upper in all terms or lower in all terms?

• no. ${}{}{}{}{}$ – AccidentalFourierTransform Jun 6 at 15:57
• The position has to match. – Lewis Miller Jun 6 at 15:57
• Would you write $\langle \psi|+|\phi\rangle$? What about $\begin{pmatrix}x_1&y_1\end{pmatrix}+\begin{pmatrix}x_2\\y_2\end{pmatrix}$? Always remember that $A^\mu$ is an element of a tangent space, $A_\mu$ of the cotangent space - its dual space! – user2723984 Jun 6 at 18:00

Well, you can write down the expression if you want to, and calculate its value, but there is a problem: it doesn't have any well defined transformation law when you change coordinates. Say you switch to a different frame through a change-of-basis matrix $$C^\mu{}_\nu$$. The components of a contravariant vector $$A^\mu$$ change to $$C^\mu{}_\nu A^\nu$$ and those of a covector $$B_\mu$$ change to $$B_\nu (C^{-1})^\nu{}_\mu$$, so if you form a combination like

$$D^\mu = A^\mu + B^\mu,$$

in a new basis we have

$$D'^\mu = C^\mu{}_\nu A^\mu + C^\mu{}_\nu B^\mu = C^\mu{}_\nu (A^\nu + B^\nu)= C^\mu{}_\nu D^\nu,\tag{1}$$

so its components will also change like a vector. But under a change of basis, the expression $$A^\mu + B_\mu$$ changes to

$$C^\mu{}_\nu A^\mu + (C^{-1})^\nu{}_\mu B_\nu$$

which does not obey any simple transformation law. In particular, and this is the important part, the way it transforms depends on the values of $$A^\mu$$ and $$B_\mu$$ individually instead of simply depending on $$A^\mu + B_\mu$$; compare with $$(1)$$ above, where the transformation just depends on $$D^\mu$$.

It's for this reason that the notation doesn't let you give it a name: both $$D^\mu = A^\mu + B_\mu$$ and $$D_\mu = A^\mu + B_\mu$$ would be wrong, because the position of the index on $$D$$ is in both cases implying a transformation law that is not the one $$D$$ actually obeys.

• Thanks for the answer, but would it be correct to write something like the following instead? $$A^{\lambda}_{\mu}+B^{\mu}_{\lambda}$$ These are two elements of a matrix for example, therefore it should be possible to write this down, or is this still incorrect? – Sørën Jun 6 at 19:12
• No, this is still not correct. You run into the same problem with the transformation matrices, which end up having the indices in the wrong positions. – Javier Jun 6 at 19:21
• Ok thanks a lot for your help! So, as a general rule, in the Einstein notation a free index must be either upper in all terms or lower in all terms of the linear combination, is this correct? – Sørën Jun 6 at 19:25
• Yes. Also, the free indices must match on both sides of an equation. – Javier Jun 6 at 19:32

When dealing with contravariant vectors (also just vectors $$\rightarrow$$ upper indices) and covariant vectors (also covectors $$\rightarrow$$ lower indices), you can only sum vectors that transform the same way with respect to change of basis.
• Thanks for the answer, but would it be correct to write something like the following instead? $$A^{\lambda}_{\mu}+B^{\mu}_{\lambda}$$ These are two elements of a matrix for example, therefore it should be possible to write this down, or is this still incorrect? – Sørën Jun 6 at 19:11
• No, it is still incorrect. You have two implicit summations in that expression, both of which are trying to sum a vector with a covector, which resides in the dual space! Something like: $$A^\lambda_\mu + B^\lambda_\mu$$ would be mathematically possible though. Can you see why? – João Bravo Jun 6 at 20:16