1
$\begingroup$

I'd like to check if my understanding of the following is correct: consider a contravariant object $x^\mu\in V$, where $V$ is a vector space with a metric $g^{\mu\nu}$. From linear algebra, we know that we can build the dual space $V^*$ by exploiting the metric: $$x^\mu\in V\to g(x^\mu,y^\nu)=x^\mu g_{\mu\nu} y^\nu\in \mathbb R $$ for some other vector $y^\nu$. So we can define a covariant object $y_\mu:= g_{\mu\nu}y^\nu$ and use it to obtain scalars by contraction with covariant vectors, as $x^\mu y_\mu$, or equivalently $x_\mu y^\mu$. In my understanding then the act of raising and lowering an index is a purely notational convention; the mathematics behind it is what I have done above with the scalar product. Is this correct? Many books don't stress this point enough in my opinion.


Thanks to everyone for your comments and answers. This is my full takeaway: please feel free to read it and correct it if necessary.

Let $v\in V$ be a vector in a four-dimensional vector space with basis $(e_\mu)$. In the index notation, we care about the components of $v$, denoted as $v^{\mu}$ for $\mu=0,1,2,3$. Sometimes, when there is no potential for confusion (although sometimes even when there is) $v^\mu$ is also taken to mean the vector $v$ itself. Explicitly, \begin{equation} v=v^\mu e_\mu=(v^0, v^1, v^2, v^3). \end{equation} A matrix $M$ acting on $V$ is also written in terms of its entries $M_{\mu\nu}$.

Now assume that $V$ has a metric, that is, a bilinear form $g: V\times V\to \mathbb R$ with the usual properties. As every linear functional can be written in terms of the metric as \begin{equation} f_w:=g(\cdot,w):V\to \mathbb R \end{equation} there is an induced isomorphism between $V$ and the dual space $V^*$: it is enough to specify which vector $w$ determines $f_w$. In index notation, consider the product \begin{equation} g(v,w):=v^T g w=v^\mu g_{\mu\nu}w^\nu \end{equation} for a fixed $w$ and any $v$, where we have introduced the Einstein's convention of summing over repeated indices. Forgetting about the '$\cdot$' part of the equation, the aforementioned isomorphism can be written in components as \begin{equation} w^\nu\to g_{\mu\nu} w^\nu:=w_\mu, \end{equation} where we have obtained an object with a lowered index (covariant vector, or covector). To be more specific, we should denote $w_\mu$ by some different letter as it lives in a different space (I have seen the use of the so called 'musical notation' $(w^\flat)_\mu=g_{\mu\nu} w^\nu$, where the flat symbol corresponds to the lowering of an index. Of course we can extend this discussion to the opposite direction, and consider the sharp objects obtained by raising an index). As the space $V^*$ is described by the (co-)basis $\alpha^\mu$ such that $\alpha^\mu(e^\nu)=\delta_\nu^\mu$, the actual covector is written as $w=w_\mu \alpha^\mu$.

When $v=w$, we have in particular \begin{equation} g(v,v)=g_{\mu\nu}v^\mu v^\nu=v_\nu v^\nu=v^2. \end{equation} If the chosen metric is the Minkowski metric $\eta_{\mu\nu}=\eta^{\mu\nu}$ with signature $(+,-,-,-)$, all of the above holds globally, as in the entire space $V=\mathbb R^4$. But in a more general manifold $M$ endowed with any metric $g$, the isomorphism is only locally defined between each tangent space $T_p V$ and its dual.

$\endgroup$
7
  • $\begingroup$ $x^\mu$ are the components of a vector in the vector space $V$, we define the dual space to be the space of all linear maps from $V$ into $\Bbb R$ (or more generally, the underlying field of $V$), we usually denote it $V^*$. The metric then provides an isomorphism between $V$ and $V^*$, so when we talk about "lowering and raising indices" what we mean is we are using the metric to associate the components of a vector with the components of the corresponding covector in the dual space under the map provided by the metric. $\endgroup$ – Charlie Apr 5 at 16:47
  • $\begingroup$ @Charlie So what I wrote is essentially correct? The thing to keep in mind is the isomorphism, and we use raised/lowered indexes just as a matter of convention (for example, repeated upper/lower indexes are to be summed over). $\endgroup$ – ForgiveMyNoobness Apr 5 at 17:04
  • $\begingroup$ On a manifold with non-trivial curvature the isomorphism is local. If you express $x^\mu$ and $x_\mu$ in coordinates, they will be different objects. The metric keeps track of how the isomorphism changes as you move around the space. $\endgroup$ – hulsey Apr 5 at 17:08
  • $\begingroup$ The association between the vector in $V$ and the corresponding covector in $V^*$ is through the inner product $\langle\cdot,\cdot\rangle$, yes. If you fill one "slot" of the inner product with a vector, $\vec v\in V$, you effectively get an object $\langle \vec v,\cdot \rangle:V\rightarrow \Bbb R$ such that $\langle \vec v,\cdot \rangle:\vec w\mapsto \langle\vec v,\vec w\rangle\in \Bbb R$. i.e. you essentially have $\langle \vec v,\cdot\rangle \in V^*$. $\endgroup$ – Charlie Apr 5 at 17:27
  • $\begingroup$ And as mentioned above there are some complications when you have a curved space in which the metric (and therefore the map between $V$ and $V^*$) is different at each point in spacetime. $\endgroup$ – Charlie Apr 5 at 17:31
2
$\begingroup$

You're in the right ball-park.

One thing that is not explained well in my opinion is the distinction between covariant and contravariant components, and as you point out, how this relies upon the metric.

If we fix a basis $e=(e_\alpha)$ in a vector space $V$, then for every vector $v \in V$, we have its components, $v^\alpha$.

So far so standard.

What's not emphasised is that in the presence of a metric we have a cobasis $f=(f_\beta)$ defined by $(e_\alpha, f_\beta)=\delta_{\alpha\beta}$. This means that we have components of $v$ with respect to this cobasis.

Hence for any vector, in the presence of a metric, we naturally have two sets of components. These are what are traditionally called contravariant and covariant components of a vector.

We may as well justifiably say the components and cocomponents with respect to the basis and cobasis.

It's only when the basis is orthogonal that the components and the cocomponents are equal and this is because in this situation, the cobasis is equal to the basis.

Notice here we didn't make any use of the dual space $V^*$. In fact, we can define a dual basis by $(e^\alpha, e_\beta)=\delta^\alpha_\beta$. Here, $e^\alpha$ is in $V^*$ and the bracket is evaluation. We can also identify linear functionals with the curryed metric and this means we identify the cobasis with the dual basis and the dual components with the cocomponents.

$\endgroup$
7
  • $\begingroup$ I think you should distinguish between inner product and covector evaluation on a vector by different notation, say $\left(e_\alpha,f_\beta\right)$ for inner product and $e^\alpha\left(e_\beta\right)$ for application. I find your notation confusing a little $\endgroup$ – Umaxo Apr 5 at 17:36
  • $\begingroup$ @Umaxo: It is standard notation in certain textbooks. The point being the bracket is generalised from an inner product $V \times V \rightarrow V$ to a bilinear product $U \times V \rightarrow W$ and then the inner product and evaluation are instances of these. The latter because evaluation is $V^* \times V \rightarrow V$. Nevertheless, I appreciate your concern - and if I was writing a textbook, I'd be more careful to distinguish and introduce the two. $\endgroup$ – Mozibur Ullah Apr 5 at 17:46
  • $\begingroup$ Sure, but they are different instances. But if you are satisfied with your notation, then of course keep it. $\endgroup$ – Umaxo Apr 5 at 18:13
  • $\begingroup$ But shouldn't it be $V \times V \rightarrow \mathbb{R}$ and $V^* \times V \rightarrow \mathbb{R}$ ? $\endgroup$ – Umaxo Apr 5 at 18:14
  • $\begingroup$ Thank you everyone for your interest. I have updated the answer with my 'takeaway' from your comments and from this answer. If any of you would like to read it and tell me if I'm getting something wrong, that would be make me happy. $\endgroup$ – ForgiveMyNoobness Apr 5 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.