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If we take a vector $A$, which has three components, my understanding is that we can write this using Einstein notation as $A_{u}$ where this is actually $A_1+A_2+A_3$. We can also write $g^{uv}A_v = A^u$.

If we take a concrete example where $A = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ and $g^{uv} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$, then if we expand $g^{uv}A_v$, we have $g^{u1}A_1 + g^{u2}A_2 + g^{u3}A_3$ which is equal to $\begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}$. My confusion is why these are both called $A$? They are two completely different vectors, so I am confused why when we contract $A_v$ with $g^{uv}$, we get another vector $A$? My understanding is that $A_u$ and $A_v$ are the same vector since they are both called $A$ and summed over $1-3$, so my intuition is that we should say $g^{uv}A_v = B^u$.

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  • $\begingroup$ You're free to use whatever notation you want: most people find Einstein's shorthand convenient when they need to look over past work however. Also, using the same letter for a vector and its (metric) dual 1-form means that you have somewhat more freedom in choosing names for other variables. It's sort of like how English and a number of other languages have ambiguities that can be resolved by context. $\endgroup$
    – TLDR
    Jun 14, 2022 at 20:20
  • $\begingroup$ You should look up covariance and contravariance of vectors and in this context the raising and lowering of indices: $g^{uv}A_v=A^u\neq A_u$. $A^u$ and $A_u$ are contravariant and covariant components of the vector A. $\endgroup$
    – N0va
    Jun 14, 2022 at 20:20
  • $\begingroup$ that was a typo not a lack of understanding. should be fixed now. @N0va $\endgroup$
    – Relativisticcucumber
    Jun 14, 2022 at 20:27
  • $\begingroup$ You still have written: "$A_\mu$ where this is actually $A_1 + A_2 + A_3$." This does not make sense. What do you mean by this? $\endgroup$
    – hft
    Jun 14, 2022 at 20:33
  • $\begingroup$ I might be wrong, but I think this might be stemming from my physics & math classes using different notations. I'm relatively certain in my physics classes $A_u$ has been used as notation for an entire vector. that is why I wrote that, because I was thinking if we have $A_u$ to represent an entire vector and we are using Einstein notation, the only way to reconcile these is to say that $A_1$, $A_2$, & $A_3$ are being added to create a full vector, but I guess I'm just confused. I guess in my head if we have [1,2,3], $A_1$ = [1,0,0], $A_2$ = [0,2,0], etc. $\endgroup$
    – Relativisticcucumber
    Jun 14, 2022 at 20:39

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If we take a vector $A$, which has three components, my understanding is that we can write this using Einstein notation as $A_{u}$ where this is actually $A_1+A_2+A_3$.

No. $A_\mu$ is the $\mu^{th}$ component of the covector $A$. Summation is only implied if an index is repeated, e.g. $A_\mu B^\mu=A_0B^0 + A_1B^1+\ldots$

We can also write $g^{\mu\nu}A_\nu = A^\mu$.

Yes. To each covector $A$ with components $A_\nu$ we can associate a vector "partner" $A^\sharp$ with components $(A^\sharp)^\mu := g^{\mu\nu} A_\nu$. It is conventional to drop the $\sharp$ and simply write $A^\mu$, where we distinguish between the components of $A$ and $A^\sharp$ by the placement of the index. Similarly, to each vector $V$ with components $V^\mu$ we can associate a covector $V^\flat$ with components $(V^\flat)_\mu := g_{\mu\nu} V^\nu$.

This can be confusing, because $A$ and $A^\sharp$ are emphatically not the same thing (they aren't even the same type of thing - one is a covector and one is a vector). They are partners, whose partnership is in this case defined by the metric tensor (though any non-degenerate bilinear form would do).

Based on some previous questions you've asked, I believe the convention used by your source lumps vectors and covectors together, treating them as elements of a single vector space. In this convention, $A^\mu$ and $A_\mu$ are called the contravariant and covariant components of the vector $A$, corresponding to the expansion of $A$ in the basis $\{\hat e_\mu\}$ or the dual basis $\{\hat \epsilon^\mu\}$, which is defined such that $\hat\epsilon^\mu \cdot \hat e_\nu = \delta^\mu_\nu$. In other words,

$$A = A^\mu \hat e_\mu = A_\nu \hat \epsilon^\nu$$

The components $A^\mu$ and $A_\mu$ are different because the bases $\{\hat e_\mu\}$ and $\{\hat \epsilon^\mu\}$ are generally different (though they coincide if $\{\hat e_\mu\}$ is an orthonormal basis).

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  • $\begingroup$ the "we can also write..." was a typo. does your comment still apply to the revised question? $\endgroup$
    – Relativisticcucumber
    Jun 14, 2022 at 20:29
  • $\begingroup$ @Relativisticcucumber Yes. See my edit at the end, though. $\endgroup$
    – J. Murray
    Jun 14, 2022 at 20:33
  • $\begingroup$ This is useful information, but I don't think it fully clarifies the question for me. So in response to $g^{\mu \nu}A_{\nu} = A^{\mu}$, you said yes. I'm still wondering why. I still feel like it should be $g^{\mu \nu}A_{\nu} = B^{\mu}$. $\endgroup$
    – Relativisticcucumber
    Jun 14, 2022 at 20:52
  • $\begingroup$ @Relativisticcucumber What is $B^\mu$? $A^\mu$ is the $\mu^{th}$ component of $A$ with respect to the basis $\{\hat e_1,\hat e_2,\ldots\}$. $A_\mu$ is the $\mu^{th}$ component of $A$ with respect to the basis $\{\hat \epsilon^1,\hat\epsilon^2,\ldots\}$, which is generally different. These components are related by $A^\mu=g^{\mu\nu}A_\nu$, which can be seen by dotting the expression $A^\mu\hat e_\mu = A_\nu\hat\epsilon^\nu$ with $\hat \epsilon^\mu$ on both sides, and noting that $g^{\mu\nu} \equiv \hat \epsilon^\mu \cdot \hat \epsilon^\nu$. $\endgroup$
    – J. Murray
    Jun 14, 2022 at 20:57
  • $\begingroup$ got it thank you very much $\endgroup$
    – Relativisticcucumber
    Jun 14, 2022 at 20:58

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