# Metric-Dual Vector Space

1. Sep 17, 2014

### ChrisVer

a metric is also used to raise/lower indices.
$g_{\nu \mu } x^{\mu} = x_{\nu}$
$g^{ \nu \mu} x_{\mu} = x^{\mu}$

In general a metric [with lower indices] is a map from $V_{(1)} \times V_{(2)} \rightarrow \mathbb{R}$
whereas the upper indices are the map from $V^{*}_{(3)} \times V^{*}_{(4)} \rightarrow \mathbb{R}$
I used the subscript to denote the vectors later I'm going to take, and the * means the dual space.
In this case then you have:
$g(x_1, x_2) = s \in \mathbb{R}$
$g(x_{3}, x_{4}) = s \in \mathbb{R}$ , with $x_{i} \in V_{(i)}^{(*)}$

Which means that $g_{\mu \nu} x_{1}^{\mu} x_{2}^{\nu} = x_{1 \nu} x_{2}^{\nu}$
and also $g^{\mu \nu} x_{3\mu} x_{4 \nu} = x^{\nu}_3 x_{4 \nu}$

Now if these are equal then it means that $g(x_{1})=g(x_4)$ or in other words the metric maps a vector in a vector space to its dual.
But isn't the dual space basis given by derivatives? So if $x^{\mu}$ is a vector, then $x_{\mu}$ should be written in terms of derivatives?

2. Sep 17, 2014

### Staff: Mentor

No; the dual space of a vector space is the space of linear maps from the vector space into the reals. So the metric maps a vector to its dual, i.e., it maps a vector into a map from vectors into reals, i.e., the covector $x_{\nu} = g_{\mu \nu} x^{\mu}$ is just the linear map that takes the vector $x^{\mu}$ into its squared length. So the inverse metric just takes a map from vectors into reals, back into a vector; i.e., the vector $x^{\mu} = g^{\mu \nu} x_{\nu}$ is the one that gets mapped to its squared length by the linear map (covector) $x_{\nu}$.

3. Sep 17, 2014

### Fredrik

Staff Emeritus
g is a map from $V\times V$ into $\mathbb R$. If $(e_\mu)_{\mu=0}^3$ is an ordered basis for V, then the 16 numbers $g_{\mu\nu}$ are called the components of g, with respect to that ordered basis. If we let $M$ denote the matrix with $g_{\mu\nu}$ on row $\mu$, column $\nu$, then $g^{\mu\nu}$ denotes the number on row $\mu$, column $\nu$ of $M^{-1}$.

Since g is a map from $V\times V$ into $\mathbb R$, it's not a map from V into V*, but you can use g to define such a map. First you note that for all x in V, the map $y\mapsto g(x,y)$ with domain V is an element of V*. It's convenient to denote this map by $g(x,\cdot)$. The map $x\mapsto g(x,\cdot)$ with domain V, is a map from V into V*.

You're using the symbol g for a lot of different things. First you use it for a metric on V, then for a metric on V* (specifically the one with components $g^{\mu\nu}$). In that final equality, I don't even know what it means. I just see that the g on the left is not the same as the g on the right (since the inputs are in different spaces).

I also don't understand what you're doing. Are you trying to determine what $x_3$ and $x_4$ would have to be in order to ensure that with this choice of metric on V*, we have $g(x_1,x_2)=g(x_3,x_4)$?

The ordered basis of V* that's dual to $(e_\mu)_{\mu=0}^3$ is the 4-tuple $(e^\mu)_{\mu=0}^3$ defined by $e^\mu(e_\nu)=\delta^\mu_\nu$ for all $\mu,\nu$.

In differential geometry, V will be the tangent space at some point p, and V* will be its dual. We can use a coordinate system x (with p in its domain) to define an ordered basis for V. For each $\mu$, we define
$$e_\mu=\frac{\partial}{\partial x^\mu}\!\bigg|_p.$$ Let me know if you don't know the definition of the right-hand side. The dual basis is defined as in the preceding paragraph. For reasons that I don't have time to go into now, the dual basis vectors can be written $\mathrm{d}x^\mu|_p$.

4. Sep 17, 2014

### dextercioby

No, you're mixing concepts. X's are coordinates on the space-time manifold. A 4-tuple of real values uniquely identify a point on the manifold (an event, or a point where one observer resides). That's why they only have 1 type of indices, namey contravariant. I hate it when people put the mu downstairs. As for the derivatives, well they form a basis of the tangent space to any point on the space-time manifold and it's easy to see that once you use the proper definition of a tangent vector.

5. Sep 17, 2014

### ChrisVer

In fact, I have a problem with this thing.
I don't understand the difference between $x^{\mu}$ and $e^{\mu}$ I think they are the same things...$x \in V$ is just a vector, so it can be written in the $e_{\mu}$ basis. And same for an $x^* \in V^*$ which can be expressed with the basis of V* $e_{*}^{\mu}$.
(http://www.itkp.uni-bonn.de/~metsch/GRC2014/grca1a.pdf)
I can't understand though how a vector can be expanded in terms of derivative basis vectors.

6. Sep 17, 2014

### DrGreg

If $\textbf{v}\in V$ you can write$$\textbf{v} = v^\mu \textbf{e}_\mu$$$v^0, v^1$, etc, are (contravariant) coordinates (i.e. numbers) and $\textbf{e}_0, \textbf{e}_1$, etc, are vectors (i.e. elements of V). Similarly if $\textbf{v}^*\in V^*$ you can write$$\textbf{v}^* = v_\mu \textbf{e}^\mu$$$v_0, v_1$, etc, are (covariant) coordinates (i.e. numbers) and $\textbf{e}^0, \textbf{e}^1$, etc, are covectors (i.e. elements of V*).

7. Sep 17, 2014

### ChrisVer

and then in order for $e^{\mu}_{*}(e_{\nu}) = \delta^{\mu}_{\nu}$ you would have to write:
$\textbf{v}^* = v_\mu \partial^{\mu}$?

8. Sep 17, 2014

### Fredrik

Staff Emeritus
You mean the same type of things? For example, if one of them is a real-valued function, then so is the other. If one of them is a 4-tuple of real numbers, then so is the other. $x$ and $e_\mu$ are the same type of things in that sense, but $x^\mu$ is a component of $x$, so it's a real number regardless of what type of object $x$ and $e_\mu$ are.

You need to know the definition of the tangent space at p to understand that. The simplest definition says that it's the vector space spanned by the set $\big\{\frac{\partial}{\partial x^\mu}\big|_p\big\}_{\mu=0}^3$. This definition makes your question trivial. A tangent vector is by definition a linear combination of the four partial derivative functionals.

There are two issues with this definition. It's ugly and it's hard to see why these things are called "tangent vectors". The ugliness issue is addressed by an equivalent definition that's coordinate independent. It goes like this: Let M be a smooth manifold. Let F be the set of smooth functions from M into $\mathbb R$. The tangent space at p is defined as the set of all linear $v:F\to\mathbb R$ such that $v(fg)=v(f)g(p)+f(p)v(g)$ for all $f,g\in F$.

The tangent space at p is denoted by $T_pM$. If we take the coordinate-independent definition of $T_pM$ as our starting point, we will have to state and prove a theorem that says that if x is a coordinate system with p in its domain, then $\big\{\frac{\partial}{\partial x^\mu}\big|_p\big\}_{\mu=0}^3$ is a basis for $T_pM$.

The other issue, that it's hard to see why these things are called tangent vectors, is best addressed by a definition of $T_pM$ that's not equivalent to the one above. This definition makes it obvious why the vectors are called tangent vectors, but is much harder to work with. So we would end up proving a theorem that says that the tangent space defined this way is isomorphic to the tangent space defined as above. And then we will end up working with the tangent space defined above anyway.

If you want to study that alternative definition of $T_pM$, check out Isham's book on differential geometry.

9. Sep 17, 2014

### Fredrik

Staff Emeritus
If $v\in V$ and $\omega\in V^*$, you can write $v=v^\mu e_\mu$ and $\omega=\omega_\mu e^\mu$. If $V=T_pM$, and $x:U\to\mathbb R^n$ is a coordinate system such that $p\in U$, then $e_\mu=\frac{\partial}{\partial x^\mu}\!\big|_p$ and $e^\mu=\mathrm dx^\mu\big|_p$. The explanation for the "d" notation is a bit complicated. You need to study 1-forms to understand it.

10. Sep 17, 2014

### ChrisVer

I meant that if $x^{\mu}$ is a vector,then so is $e^{\mu}$. The only difference is the name, and that e is some basis (like $e_{x,y,z}$ is a basis for the 3dimensional euclidean space).

I guess the tangent space is easier to see in a 2dim surface which is the mapping $x: U_2 \rightarrow S$ where $U$ is the 2d (u,v) parameter space and $S$ is the surface, where then the tangent space is spanned by the derivatives of x wrt the parameters of your surphase... So $d \textbf{x}(u,v) = \textbf{x}_u du + \textbf{x}_v dv$ right? so $\textbf{x}_{u,v}$ are the vectors which span the tangent plane of the surface. Now on some point p, then you have to take the derivatives at the corresponding point $\textbf{x}_{u,v} (u_0,v_0)$ if p is in $(u_0,v_0)$.

Now going to a 4dimensional case, I have $\textbf{X}(x^{\nu})$,_where now $\textbf{X}: U_{4} \rightarrow M$ maps the $x^{\mu} \in U_4$ coords to some manifold M, and thus in the same way:
$d \textbf{X} (x^{\mu})= \partial_{\mu} \textbf{X} dx^{\mu}$
and thus the spanning vectors are the $\partial_{\mu} \textbf{X}$

Last edited: Sep 17, 2014
11. Sep 17, 2014

### Staff: Mentor

It isn't. It's a component of a vector. If $\vec{e}_{\mu}$ (note that the index is lower here, not upper) is a vector--a basis vector--then the vector $\vec{x}$ can be written in component form as $x^{\mu} \vec{e}_{\mu}$. In other words, the vector itself is a linear combination of the basis vectors; the components of that vector are just the coefficients in the linear combination.

I agree that this notation for components and vectors can be confusing (the index $\mu$ labels both components and basis vectors). But that's the standard notation.

Now you're confusing the notation: is $x$ supposed to be a vector? Vectors aren't mappings; they're elements of a vector space. Covectors are mappings, but they're mappings of vectors into numbers, not mappings of pairs (or n-tuples) of numbers into points in a surface. The latter type of mapping is a coordinate chart, though properly speaking it should be given the other way around: a coordinate chart is a mapping of points in a manifold into n-tuples of real numbers, where n is the dimension of the manifold.

Perhaps what you're trying to say here is that the *components* of a vector $x$ can be viewed as n-tuples of numbers, and therefore, if we choose some particular point in the surface as the origin, we can view the components of a vector $x$ as the n-tuple that the coordinate chart maps a given point to, where the point is the one at the tip of the vector when its tail is at the origin. However, you should be aware that this doesn't work in a curved manifold, only in a flat one.

Strictly speaking, vectors do not "live" in the manifold itself; they live in the tangent space at a particular point (there is a different tangent space at each point); but in a flat manifold, you can get away with identifying the tangent space at any given point with the manifold itself, described in a chart where the given point is the origin. This is a bad habit to get into, though, because it doesn't generalize to curved manifolds. It's better to keep vectors, points, and coordinate n-tuples all separate in your mind, and to keep in mind also that all of these objects "live" in the tangent space at a given point in the manifold, not in the manifold itself.

No, not "derivatives of x". First of all, that just repeats the above confusion: is $x$ supposed to be a vector? A point? An n-tuple of coordinates? Second, the derivatives that span the tangent space are not derivatives "of" anything in particular; they are directional derivative operators along particular orthogonal directions--more precisely, they are directional derivatives along curves in particular orthogonal directions, evaluated at the given point. The reason this works is that there is a bijective correspondence between tangent vectors at a given point, and directional derivative operators along curves at that point: the operator corresponding to a given tangent vector is just the directional derivative along the curve that the vector is tangent to. So the derivatives that span the tangent space are just the directional derivatives corresponding to the basis vectors.

Last edited: Sep 18, 2014
12. Sep 18, 2014

### Fredrik

Staff Emeritus
$x^\mu$ would be an awful notation for a vector. It typically denotes a component of a vector $x=x^\mu e_\mu$, or a component of a coordinate system. If $x$ and $e_\mu$ are 4-tuples of real numbers, then $x^\mu$ is a real number. But if $x:U\to\mathbb R^n$ is a coordinate system on a manifold and $p\in U$, we would write $x(p)=(x^1(p),x^1(p),x^2(p),x^n(p))$. (The default choice for indices is to let them go from 1 to n, but in relativity we let them go from 0 to 3 instead of 1 to 4). So we would also use the notation $x^\mu$ for the map that takes a point p in U to it's $\mu$th coordinate in the coordinate system x.

In this case it would make sense to define the tangent space at $(a,b)\in U_2$ as derivatives of curves in S through the point (a,b). The maps $u\mapsto \mathbf x(u,b)$ and $v\mapsto\mathbf x(a,v)$ are curves in S through $\mathbf x(a,b)$. Their derivatives at the points in their domains that are mapped to $\mathbf x(a,b)$ are $\mathbf x_u(a,b)$ and $\mathbf x_v(a,b)$, where $\mathbf x_u$ and $\mathbf x_v$ denote the partial derivatives of $\mathbf x$. $\mathbf x_u(a,b)$ and $\mathbf x_v(a,b)$ are elements of $\mathbb R^3$. It makes sense to think of the plane spanned by these vectors as the tangent plane of the surface $\mathbf x$ at the point $\mathbf x(a,b)$.

But this is not how it's done in differential geometry. If $C:\mathbb R\to M$ is a curve in a manifold M, you can't just compute $C'(0)$ and call it a tangent vector, because there's no addition operation on M. But if x is a coordinate system, then $x\circ C$ is a curve in $\mathbb R^n$. The components of the tangent vector of C at the point C(0), in the coordinate system x, are $(x^\mu\circ C)'(0)$. The tangent vector can be written as $(x^\mu\circ C)'(0)e_\mu$, but the $e_\mu$ in this equality are not the standard basis vectors for $\mathbb R^n$. They're the partial derivative functionals $\frac{\partial}{\partial x^\mu}\!\big|_{C(0)}$.

13. Sep 20, 2014

### stevendaryl

Staff Emeritus
Well, there is a convention--which I don't particularly like, although lots of people follow it--that a variable with a superscript is interpreted as a vector, and a variable with a subscript is interpreted as a covector. So when someone following this convention writes $v^\mu$, they mean what you would write as: $v^\mu e_\mu$.