# Notation for contracting vectors using metric tensors

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 $$g^{uv}A_v = B^u$$.

• 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.
– TLDR
Jun 14, 2022 at 20:20
• 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.
– N0va
Jun 14, 2022 at 20:20
• that was a typo not a lack of understanding. should be fixed now. @N0va Jun 14, 2022 at 20:27
• 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?
– hft
Jun 14, 2022 at 20:33
• 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. Jun 14, 2022 at 20:39

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).

• the "we can also write..." was a typo. does your comment still apply to the revised question? Jun 14, 2022 at 20:29
• @Relativisticcucumber Yes. See my edit at the end, though. Jun 14, 2022 at 20:33
• 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 $g^{\mu \nu}A_{\nu} = B^{\mu}$. Jun 14, 2022 at 20:52
• @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 $g^{\mu\nu} \equiv \hat \epsilon^\mu \cdot \hat \epsilon^\nu$. Jun 14, 2022 at 20:57
• got it thank you very much Jun 14, 2022 at 20:58