# Position of Indices in Einstein's Summation Convention

Would there be a difference between the following tensor quantities?

$${A_\gamma}^\mu {B_\mu}^\rho$$ and $${A^\mu}_\gamma {B_\mu}^\rho$$

Would both of these expressions give the same result as follows $${X_\gamma}^\rho = {A_\gamma}^\mu {B_\mu}^\rho = {A^\mu}_\gamma {B_\mu}^\rho?$$

As the positions of the indices in Einstein's summation convention make a difference, I believe that the two expressions are not equivalent. The first equation ${X_\gamma}^\rho = {A_\gamma}^\mu {B_\mu}^\rho$ makes sense to me as the positions of the indices $\gamma$ and $\rho$ are clear in the resultant quantity ${X_\gamma}^\rho$ but I am unsure of the second equation. Would ${X_\gamma}^\rho = {A^\mu}_\gamma {B_\mu}^\rho$?

Would there be a difference between the following tensor quantities?

Yes. Your equation is false. The l.h.s. contracts the second index, and the r.h.s. one contracts the first one (even if you are using the same letter for the index). Unless your tensors have some sort of symmetry, these expressions do not agree.

• It would be helpful to add that if you make an analogy with matrix multiplication, it is like comparing $\mathbf{AB}$ and $\mathbf{A}^T\mathbf{B}$. – Brian Moths Dec 16 '16 at 19:57
• Would ${X_\gamma}^\rho = {A_\gamma}^\mu {B_\mu}^\rho$ and $Y_{\hspace{6pt}\gamma}^{\hspace{3pt}\rho} = {A^\mu}_\gamma {B_\mu}^\rho$ where $\rho$ and $\gamma$ are in the second index positions for $Y_{\hspace{6pt}\gamma}^{\hspace{3pt}\rho}$? – Charles Dec 18 '16 at 12:36
• @Charles what are you trying to write? note that $A^\mu{}_\nu$ can be written as A^\mu{}_\nu and $A_\mu{}^\nu$ can we written as A_\mu{}^\nu – AccidentalFourierTransform Dec 18 '16 at 12:38
• Would ${A^\mu}_\gamma {B_\mu}^\rho$ contract over the first index to give a tensor $Y_{\hspace{6pt}\gamma}^{\hspace{3pt}\rho}$ where $\rho$ and $\gamma$ are in the second index positions? Or would having two indices in the same positions be an invalid notation? – Charles Dec 18 '16 at 12:43
• In $I^\rho_\gamma$ the indices are not in the same position: one is an upstairs and one is downstairs. If you are worried about the horizontal position of the index, I fear that there is little consensus: for some people it is right to write $I^\mu{}_\nu$ simply as $I^\mu_\nu$, ignoring the horizontal position. For some other people such a notation can only be used if $I$ is symmetric. And for some people, you cannot write $I^\mu{}_\nu$ as $I^\mu_\nu$. In the end, as long as you understand what you are writing, and are consistent and clear with the notation, you can write whatever you want. – AccidentalFourierTransform Dec 18 '16 at 12:49

Let's start with the metric $\eta_{\mu\nu}$ such that $\Delta s^2 = \Delta x^\mu \; \eta_{\mu\nu} \; \Delta x^\nu$. Denote the inverse metric as $\eta^{\mu\nu}$ so that $\eta_{\mu\nu}\eta^{\nu \rho} = \delta_{\mu}^{\,\,\rho}$.

Start with tensor with only index up, like $$A^{\gamma\mu} \qquad \qquad \text{and} \qquad \qquad B^{\mu \rho}$$ then define the lower-index version of these tensors by contracting with the metric $\eta^{\mu\nu}$: $$A_{\gamma}^{\;\,\mu} := \eta_{\gamma \alpha} A^{\alpha \mu}; \qquad \qquad A_{\,\;\mu}^{\gamma} := \eta_{\mu \alpha} A^{ \gamma\alpha}$$ and so on.

You see that for instance $$X_{\gamma}^{\;\;\rho} = A_{\gamma}^{\;\;\mu} B^{\;\;\rho}_{\mu} \equiv \eta_{\gamma \alpha} \eta_{\mu\beta} A^{\alpha\mu}B^{\beta\rho}$$ whereas $$Y_{\gamma}^{\;\rho} = A_{\;\;\gamma}^{\mu} B^{\;\;\rho}_{\mu} \equiv \eta_{\gamma \alpha} \eta_{\mu\beta} A^{\mu\alpha}B^{\beta\rho}$$ and you see that are, in general, two different tensors, and only if $A$ (in this case) is symmetric they are equal.

But note that calling the first quantity $X_{\gamma}^{\;\;\rho}$ or$X_{\;\;\gamma}^{\rho}$ is only a matter of definition, they would both be consistent with index conventions.