# Why do we select the metric tensor for raising and lowering indices?

Any rank 2 tensor would do the same trick, right?

Then, which is the motivation for choosing the metric one?

Also, if you help me to prove that $g^{kp}g_{ip}=\delta^k_i$, I would be thankful too ^^

• i) You need a covariant 2-tensor that induces a nondegenerate bilinear form on each tangent space. ii) The tensor with components $g^{ij}$ is by definition the tensor satisfying $g^{ij}g_{jl}=\delta^i{}_l$, i.e. $g^{-1}g=I$ holds in the sense of matrices. – Ryan Unger May 21 '17 at 0:24

Every vector space $V$ (over the real numbers) comes naturally with it's dual vector space $W$ of linear maps to the real numbers, e.g. $$\varphi \in W :\Leftrightarrow \varphi: V \rightarrow \mathbb{R} ~ \text{ with } ~ \varphi(a \cdot v + u) = a \cdot \varphi(v) + \varphi(u) ~ ~ \forall v,u \in V ~ ~ \forall a \in \mathbb R$$

Let $B := \{e_1, e_2, \dots, e_n \}$ be a basis of $V$ then for $x \in V$ the expansion $$x = \sum_{i=1}^n X^i e_i$$ yields the so called contravariant coordinates $X^i$ of $x$.

The dual basis to $B$ is defined as the basis $\tilde{B} = \{ \varepsilon^1, \varepsilon^2, \dots, \varepsilon^n \}$ of $W$ such that $$\varepsilon^i(e_j) = \delta^i_j$$ which defines $\tilde{B}$ completely because every linear map is already fixed if it is given for all basis vectors of $V$. Then for a $y \in W$ the expansion $$y = \sum_{i=1}^n Y_i \varepsilon^i$$ yields the so called covariant coordinates $Y_i$ of $y$.

If on the vector space $V$ there is an inner product $(~.~,~.~): V \times V \rightarrow \mathbb{R}$ (for instance the usual dot product) then there is a natural map $\tau: V \rightarrow W$ between $V$ and $W$:

Let $x \in V$ then $$\tau(x) = (x,~.~) \in W$$ e.g. for $v \in V$ we have $$\tau(x)(v) = (x,v)$$

Let's define the following matrix $g$ $$g_{i j} := (e_i, e_j)$$ which are the (covariant) components of the twice linear inner product map, also called the metric tensor. In general the basis $B$ is not orthonormal, that is, in general $g_{i j} \ne \delta_{ij}$ and therefore $\tau(e_i) \ne \varepsilon^i$

Define the inverse matrix of $g_{ij}$ by $$g^{ij} := (g^{-1})^{ij}$$ then we have $$\sum_{k=1}^n g^{ik} \tau(e_k)(e_j) = \sum_{k=1}^n g^{ik} (e_k,e_j) = \sum_{k=1}^n g^{ik} g_{kj} = \delta^i_j$$

So indeed we have found: $$\varepsilon^i = \sum_{k=1}^n g^{ik} \tau(e_k)$$

Let's have a quick recap what we have done so far.

1. Every (real) vector space $V$ has a dual vector space $W$ attached to it
2. The inner product $(~.~,~.~)$ defines a natural map $\tau: V \rightarrow W$
3. The metric tensor $g_{ij}$ can be used to map basis vectors $e_k$ to dual basis vectors $\varepsilon^i$

Now we have everything at hand to express the covariant (dual) coordinates $X_i$ in terms of the contravariant coordinates $X^i$ of a vector $x \in V$. Remember: $$x = \sum_{i=1}^n X^i e_i$$ Now map it to the dual vector $\tau(x)$ which is a representation of $x$ in the dual vector space $W$: $$\begin{split} \tau(x) & = (x, ~.~) = (\sum_{i=1}^n X^i e_i, ~.~) = \sum_{i=1}^n X^i (e_i, ~.~) \\ & = \sum_{i=1}^n X^i \tau(e_i) = \sum_{i=1}^n \sum_{j=1}^n X^i \delta^j_i \tau(e_j) \\ & = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n X^i g_{ik} g^{kj} \tau(e_j) \\ & = \sum_{k=1}^n \left( \sum_{i=1}^n X^i g_{ik} \right) \left( \sum_{j=1}^n g^{kj} \tau(e_j) \right) \\ & = \sum_{k=1}^n \underbrace{\left( \sum_{i=1}^n X^i g_{ik} \right)}_{=X_k} ~ \varepsilon^k \end{split}$$

And there you have it $$X_k = \sum_{i=1}^n X^i g_{ik}$$ the covariant coordinates are given by the contraction of $X^i$ and $g_{ik}$. One can now go ahead and show that the natural map $\tau$ is invertible and proof the similar result $$X^i = \sum_{k=1}^n g^{ik} X_k$$ but this should be clear since $g_{ij}$ is invertible (if it wasn't, $B$ would be no basis because some $e_i$ were linear combinations of the others).

The crucial thing to observe is that $X^i$ and $X_i$ are not some random numbers but two different things: The former being the coordinates of $x$ in $V$ while the later being the coordinates of $\tau(x)$ in $W$.

• Do note: this is for finite dimensional vector spaces only – image May 24 '17 at 13:56
• What do you use the property that the metric tensor is non-degenerate and symmetric? Also, how do you know that the inverse of the metric tensor exists? – Shuheng Zheng May 19 at 4:10
• The metric tensor is just the coordinate expression of the inner product (see definition of g), which is by definition symmetric and non-degenerate over a real vector space: en.wikipedia.org/wiki/Inner_product_space – image May 19 at 21:58

To have a good notion of "raising and lowering indices", you need to have a non-degenerate 2-tensor. If it is not non-degenerate, you might send $v^\mu$ to $v_\mu=0$, in which case you can't say that "lowering the index and then raising it back up" is like doing nothing. Being able to do that is of vital importance. A Riemannian manifold comes, by definition, equipped with such a non-degenerate (symmetric) bilinear form, so there is a canonical choice of notion of raising and lowering indices.

If your manifold is guaranteed to carry other non-degenerate 2-tensors, then you may use those as well to raise and lower indices (though the meaning is different from that of raising and lowering with the metric). This is the case, for instance, if your manifold is symplectic (i.e. carries a closed, non-degenerate 2-form $\omega$).

• +1 Non-degeneracy is key. But I think it would also be worth pointing out that physics has some things to say about any particular choice. For example, the signature of the metric is important to understanding the meaning of the norm of a vector that comes from that metric. And the scaling allows us to relate the norm to particular values of other physical quantities. – Mike May 17 '17 at 12:52

It is a convention$$^1$$. If a theory has a distinguished invertible rank-2 tensor, why not use it? Examples:

1. The metric $$g_{\mu\nu}$$ in GR.
2. The $$\epsilon_{\alpha\beta}$$ metric to raise and lower Weyl spinor indices.
3. The symplectic 2-form $$\omega_{IJ}$$ in symplectic geometry.

If there are more than one distinguished invertible rank-2 tensor, one would have to make a choice. E.g. bi-metric GR, etc.

--

$$^1$$More formally, the notion of "raising and lowering indices" reflects the musical isomorphism. See also e.g. this Phys.SE post and links therein.

The metric tensor is defined by its ability to raise and lower indices. Take a finite dimensional vectors space, $V$, with $\operatorname{dim}\{V\} = N$. Given another space, $W$, with dimension $M$ you can construct the space of linear maps between those spaces. We call elements from the space of linear maps matrices, and in this example, they form a vector space that has $\operatorname{dim}\{L : V\rightarrow W\} = N\times M$.

If the target space has dimension $1$ (i.e. the map from $V$ to scalars) then the space of maps has dimensions $N$. As we learned from linear algebra, any two finite dimensional spaces with the same dimensionality are isomorphic. We can therefore construct an isomorphic map (i.e. a map that is both 1 to 1 and covers the target space) from the $V$ to $L$. The map from $V$ to $L$ is called the metric.

In other words, when the index is up, the vector is from $V$, when down it's from $L$ (often called the dual-space).

Invertability aside, we also pick the metric to have other properties we desire. In particular, we want the scalars produced by $g(v_1) v_2$ to be invariant under some set of transformations (usually rotations or Lorentz transformations). This is what constrains the metric signature (pattern of signs of eigenvalues).

Also, there is no proof that $g^{\mu \nu} g_{\nu \alpha} = \delta^\mu_{\hphantom{\mu}\alpha}$ because that is the definition of the inverse metric (the map from $L$ to $V$, as opposed to the other way around).

Using the metric tensor in this way accounts for the Riesz representation theorem in a non-orthogonal basis.

The Riesz representation theorem establishes a correspondence between linear functionals and vectors by associating a linear functional with the vector that is normal to it's level sets. Contracting with the metric tensor computes this normal vector.

What it means for a vector to be "normal" to a hyperplane depends on the metric, so if you replace the metric tensor with another bilinear form, you will be computing the vector that is "perpendicular" in a different metric.

Another way of looking at this is with $1$-forms and vectors. Given the vector basis $\frac{\partial}{\partial x^a}$ and the $1$-form ${\bf\omega}^b$ these are dual if $${\bf\omega}^b\left(\frac{\partial}{\partial x^a}\right)~=~\delta_a^b.$$ The metric as it changes covariant indices to contravariat indices, to use a somewhat older definition, means that with vector $X_a$ that $$g_{ab}~=~g(X_a,~X_b),$$ and in a dual meaning that $$g^{ab}~=~g({\bf\omega}^a\otimes{\bf\omega}^b),$$ here as a symmetric product. The role of the metric tensor in raising and lowering indices is tied to the dualism between vectors and forms.

I got this one:

We know that we can write a vector as: $$\vec{A}=A^ie_i=A_ie^i$$

Then we expect: $$\vec{A}\cdot e_i=A_i$$ $$\vec{A}\cdot e^i=A^i$$

Which leads to: $$A_i=\vec{A}\cdot e_i=(A^ke_k)\,e_i=A^k(e_k\,e_i)$$

But $e_k\,e_i$ is our natural definition of $g_{k,i}$.

So: $$A_i=A^kg_{ki}$$

## protected by Qmechanic♦May 17 '17 at 19:28

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