Confusion about upper and lower indices

Question

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1. What the student is asking

The student wants to compute the components of the product

[ X^{T}\,g , ]

where

(X) is a (1,1)‑type tensor (a matrix) with components (X^{\mu}{}_{\nu});
(g) is the metric tensor with components (g_{\mu\nu}=g_{\nu\mu}) (or its inverse (g^{\mu\nu}));
The superscript T denotes the matrix transpose, i.e. ((X^{T})^{\mu}{}{\nu}=X{\nu}{}^{\mu}) when the metric is used to raise/lower the indices.

The student tried two different ways of “expanding” the product, obtained

[ X_{\nu}{}^{\mu}\qquad\text{and}\qquad X^{\nu}{}_{\mu}, ]

and wonders which step is illegal and why the two results differ.

2. Detailed, step‑by‑step solution

2.1 Conventions we will use

Symbol	Meaning
(X^{\mu}{}_{\nu})	components of a (1,1) tensor (a linear map)
(g_{\mu\nu})	metric (covariant) – lowers an index: (V_{\mu}=g_{\mu\nu}V^{\nu})
(g^{\mu\nu})	inverse metric – raises an index: (V^{\mu}=g^{\mu\nu}V_{\nu})
(\delta^{\mu}{}_{\nu})	Kronecker delta, the identity matrix
((X^{T}))	matrix transpose with respect to the metric: ((X^{T})^{\mu}{}{\nu}=X{\nu}{}^{\mu}) . In index‑free language, the transpose is the adjoint defined by (g(Xu,v)=g(u,X^{T}v)).

Important: Raising or lowering an index must be done with the metric, never by simply swapping the position of an index. In other words, (X^{\mu\nu}\neq X^{\nu\mu}) unless the tensor is already symmetric; you have to insert a metric (or its inverse) to move an index from up to down.

2.2 Write the product with explicit index contractions

The matrix product ((X^{T}g)^{\mu}{}_{\nu}) means

[ (X^{T}g)^{\mu}{}{\nu}= (X^{T})^{\mu}{}{\rho}\,g^{\rho}{}_{\nu}\; , ]

or, equivalently (because the metric can be written with both indices down),

[ (X^{T}g)^{\mu}{}{\nu}= (X^{T})^{\mu\rho}\,g{\rho\nu}. ]

Both forms are correct; they are related by (g^{\rho}{}{\nu}=g^{\rho\lambda}g{\lambda\nu}).
The key point: the index that is summed over (the dummy index) must appear once up and once down in each factor.

2.3 Insert the definition of the transpose

By definition of the transpose with respect to the metric,

[ (X^{T})^{\mu}{}{\rho}=X{\rho}{}^{\mu}=g_{\rho\alpha}X^{\alpha\mu}\; . ]

If we prefer the version with both indices up,

[ (X^{T})^{\mu\rho}=X^{\rho\mu}=g^{\rho\alpha}X_{\alpha}{}^{\mu}. ]

Both are legitimate; they just use different positions for the metric that performs the index‑raising/lowering.

2.4 Compute the product using the first convenient form

Take

[ (X^{T}g)^{\mu}{}{\nu}= (X^{T})^{\mu\rho}\,g{\rho\nu}. ]

Insert ((X^{T})^{\mu\rho}=X^{\rho\mu}):

[ \begin{aligned} (X^{T}g)^{\mu}{}{\nu} &= X^{\rho\mu}\,g{\rho\nu}\[2pt] &= \bigl(g^{\rho\alpha}X_{\alpha}{}^{\mu}\bigr) g_{\rho\nu}\[2pt] &= X_{\alpha}{}^{\mu}\,\underbrace{g^{\rho\alpha}g_{\rho\nu}}{\delta^{\alpha}{}{\nu}}\[2pt] &= X_{\nu}{}^{\mu}. \end{aligned} ]

Result 1:

[ \boxed{(X^{T}g)^{\mu}{}{\nu}=X{\nu}{}^{\mu}}. ]

2.5 Compute the product using the second convenient form

Now start from

[ (X^{T}g)^{\mu}{}{\nu}= (X^{T})^{\mu}{}{\rho}\,g^{\rho}{}_{\nu}. ]

Insert the transpose with one index down:

[ \begin{aligned} (X^{T}g)^{\mu}{}{\nu} &= X{\rho}{}^{\mu}\,g^{\rho}{}{\nu}\[2pt] &= X{\rho}{}^{\mu}\,g^{\rho\lambda}g_{\lambda\nu}\[2pt] &= \underbrace{g^{\rho\lambda}X_{\rho}{}^{\mu}}{X^{\lambda\mu}}\,g{\lambda\nu}\[2pt] &= X^{\lambda\mu}g_{\lambda\nu}\[2pt] &= X_{\nu}{}^{\mu}. \end{aligned} ]

Again we obtain (X_{\nu}{}^{\mu}).
The step where the student wrote

[ X^{\rho}{}{\mu}\,\delta^{\rho}{}{\nu}=X^{\rho}{}{\mu}\,\delta^{\nu}{}{\rho}=X^{\nu}{}_{\mu} ]

is illegal because the Kronecker delta (\delta^{\rho}{}{\nu}) can only be used to replace the dummy index (\rho) by (\nu) when the index positions match (upper‑to‑lower). Changing the order of the delta as (\delta^{\nu}{}{\rho}) swaps the positions of the free indices, which is not allowed without an accompanying metric.

2.6 Where the student made the mistake

Let us pinpoint the exact illegal step in the second chain:

[ \begin{aligned} (X^{T}g)^{\mu}{}{\nu} &= (X^{T})^{\mu}{}{\rho} g^{\rho}{}{\nu}
&= X^{\rho}{}{\mu} g^{\rho}{}{\nu}\quad\text{(transpose definition)}
&= X^{\rho}{}{\mu}\,\delta^{\rho}{}{\nu}\qquad\text{(since }g^{\rho}{}{\nu}= \delta^{\rho}{}{\nu}\text{)}
&= X^{\rho}{}{\mu}\,\delta^{\nu}{}{\rho}\quad\color{red}{\text{illegal}}
&= X^{\nu}{}{\mu}\; . \end{aligned} ]

The equality (\delta^{\rho}{}{\nu}= \delta^{\nu}{}{\rho}) holds, but you cannot then replace the dummy index (\rho) by the free index (\nu) in the opposite position. The dummy index must be eliminated exactly as it appears in the delta:

[ X^{\rho}{}{\mu}\,\delta^{\rho}{}{\nu}=X^{\nu}{}_{\mu}\quad\text{(correct)}. ]

If you rewrite the delta as (\delta^{\nu}{}{\rho}), the summed index is now (\rho) in the lower position, so the contraction would give (X^{\nu}{}{\mu}) only if the free index (\nu) were lower, i.e. (X_{\nu\mu}). Because the index positions are mismatched, the step is not allowed.

In short: Never move a free index from an upper to a lower (or vice‑versa) position without inserting the metric. The Kronecker delta alone cannot change the variance of an index.

3. Final answer

Using the proper definition of the transpose and the rule that a dummy index must appear once up and once down in each factor, the product of the transpose of (X) with the metric is

[ \boxed{(X^{T}g)^{\mu}{}{\nu}=X{\nu}{}^{\mu}} . ]

Both of the student’s expansion routes give the same result provided the index‑raising/lowering is performed correctly. The erroneous step was the replacement

[ X^{\rho}{}{\mu}\,\delta^{\rho}{}{\nu}\;\longrightarrow\; X^{\rho}{}{\mu}\,\delta^{\nu}{}{\rho}, ]

which swaps the position of the free index without a metric and therefore changes the variance of the index illegally.

4. Common Mistakes in this type of problem

Mistake	Why it is wrong	How to avoid it
Swapping the position of a free index inside a Kronecker delta (e.g. (\delta^{\rho}{}{\nu}\to\delta^{\nu}{}{\rho})) and then using it to replace the dummy index.	The delta only identifies the same index with the same variance. Changing its order changes the variance of the free index, which is not allowed.	Keep the delta exactly as it appears; perform the contraction directly: (A^{\rho} \delta_{\rho}^{\ \nu}=A^{\nu}).
Treating raising/lowering as a mere “move” of the index (writing (X^{\mu\nu}=X^{\nu\mu}) or (X_{\mu}{}^{\nu}=X^{\nu}{}_{\mu})).	Raising/lowering requires the metric: (X_{\mu}{}^{\nu}=g_{\mu\alpha}X^{\alpha\nu}). Without the metric the equality is not guaranteed.	Whenever you need to change the position of an index, insert (g_{\alpha\beta}) or (g^{\alpha\beta}) explicitly.
Using the metric as the identity matrix without checking index positions (writing (g^{\rho}{}{\nu}= \delta^{\rho}{}{\nu}) and then treating the delta as if it had opposite variance).	The identity property holds only when the metric contracts a raised with a lowered index.	Remember: (g^{\rho}{}{\nu}= \delta^{\rho}{}{\nu}) only when the first index is up and the second is down.
Confusing the transpose with the simple index swap ((X^{T})^{\mu}{}_{\nu}=X^{\nu\mu}).	The transpose is defined via the metric: ((X^{T})^{\mu}{}{\nu}=X{\nu}{}^{\mu}=g_{\nu\alpha}X^{\alpha\mu}).	Write the definition of the transpose explicitly in terms of the metric before manipulating components.
Leaving a dummy index appearing twice in the same position (e.g. (X^{\mu}{}_{\mu}) without a sum).	Dummy indices must be summed; they cannot appear twice in the same variance position because the Einstein summation convention requires one up and one down.	Use distinct dummy letters, e.g. (X^{\mu}{}_{\nu}\,g^{\nu\rho}).

By respecting these conventions, index gymnastics become reliable and free of contradictions.

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