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I am watching MIT open course lectures on GR https://www.youtube.com/watch?v=H6eR3sG524M&t=3614s

The lecturer introduces raising and lowering indices around 1 hour mark in the lecture.

When I try to read about raising and lowering indices online, the sources often talk about bilinear forms and covariant / contrvariant vectors. I don't have a grasp of those concepts. However, in this lecture course specifically, the lecturer said those things won't be needed for the course i.e. the course is not assuming the knowledge of this.

I can also think of lower and upper indices as rows and columns of a matrix. But the lecturer in the video is saying one should not think of the algebra shown in the lecture as matrix operations with columns and rows and treat it in a more abstract way.

I don't think the lecture actually introduces what upper and lower indices mean. He does introduce one forms. Am I right in my understanding that a one-form is just a linear map of a vector to scalar?

My questions are:

  • What's the difference between lower and upper index of a tensor?
  • What is the meaning of raising and lowering indices?
  • Which part of it is notation / convention and which part has actual mathematical and physical meaning?
  • Why does multiplying a tensor by a metric lower / raise index? What's the proof?

I would appreciate a more simplistic explanation, I am studying GR as a hobby my understanding of the math is not 100% rigorous.

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    $\begingroup$ The horizontal ordering left-right of indices of a two-index tensor is what determines it is a row or a column when you write it as a matrix. Up v.s. down tells us about contravariance covariance, which means something completely else. Vectors have upper indices and one-forms have lower indices in physics, because we mean the components when we write that stuff down. Mathematicians have the opposite convention. This is one of those things that are horribly taught. $\endgroup$ Aug 17, 2023 at 15:14
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    $\begingroup$ So, to understand the raising and lowering of indices I need to have a grasp of covariance and contravariance? And I was misled by the lecture saying I can proceed without a grasp of those concepts? $\endgroup$
    – Ilya Lapan
    Aug 17, 2023 at 16:10
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    $\begingroup$ Both are correct. To understand raising and lowering, you must understand covariance and contravariance; they are actually the same thing and it will become superbly natural if you understand them. It is a fact of life, a necessary complication. But the lecturer can also be correct that, in his course, you don't have to care. He can be writing in the correct notation for future use but not gonna bother explaining from within the course. $\endgroup$ Aug 17, 2023 at 17:40
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    $\begingroup$ Two points. (1.) I think you do need to understand bilinear forms, and contravariant/covariant vectors, in order to really understand what upper and lower indices mean. You don't need to understand those concepts to do calculations so long as you're willing to take the rules "as given," but I don't think you'll have a good understanding of why the rules are the way they are. (2.) Why do you expect this topic to have a "physical intuition?" I think this is an intrinsically mathematical subject. Tensors are a tool used to describe physical systems, they aren't physical in themselves. $\endgroup$
    – Andrew
    Aug 17, 2023 at 18:22
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    $\begingroup$ The lecturer simply lied if he said you won't need knowledge of bilinear forms. The fundamental object of study in GR is the metric, which is by definition precisely a symmetric, non-degenerate bilinear form on the space-time manifold (usually taken to be of signature $(3,1)$). To say you can study GR without some understanding of bilinear forms is like saying you can study Newtonian Mechanics without a working knowledge of vectors. $\endgroup$ Aug 18, 2023 at 2:38

6 Answers 6

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A component-centric study of tensors can be simultaneously very useful and rather opaque. The deal of raising and lowering indices and even the difference between them can become rather artificial and confusing. From what I can gather from the video you linked, you learned about tensors for what they are, as multilinear maps, but even so I will give a brief exposition to answer your first question

What's the difference between lower and upper index of a tensor?

So, what is a tensor? First, given a vector space $V$, one can construct its dual space, the space of all linear maps from $V$ to $\mathbb{R}$, $V^*$. Then, a tensor on $V$ is a map $$T: \underbrace{V\times...\times V}_{\text{p times}}\times \underbrace{V^*\times ...\times V^*}_{\text{q times}}\rightarrow \mathbb{R}$$ which is linear in every one of its arguments in the obvious way. We classify tensors by how many times they fetch members of the vector space and its dual: we call $T$ a $(p,q)$-$\text{tensor}$.

Now, reconsider $V$ and $V^*$. We call the members of $V$ vectors and those of $V^*$ covectors. Just like on $V$ you can find a basis and write any vector in terms of some components, on $V^*$ you can find a basis and write any covector in terms of components. Say we found $\{e_k\}$ and $\{\epsilon_k\}$ as some basis of $V$ and $V^*$. Then, we can also expand $T$ on components \begin{align*}T(v_1,...v_p,\omega_1,...\omega_q)&=T\left(\sum_i v_i^1 e_i,...,\sum_j \omega_j^1 \epsilon_j,...\right) \\ &=\sum_i\sum_k... v_i^1 ...v_l^p \omega_j^1 ...\omega_m^q T(e_i,...,e_l,\epsilon_j,...,\epsilon_m) \end{align*} where the "..." mean that I'm taking $p+q$ sums of products of $p+q$ components and the result of $T$ in each of the basis vectors and covectors; essentially, I expanded everything in a basis and then used the multi-linearity of the tensors to "pull everything out". The $$T(e_i,...,e_l,\epsilon_j,...,\epsilon_m)$$ are called the components of the tensor. Now you'll notice that I'm not yet making any distinction about lower and upper indices. But I'm going to make it now, and hopefully it will make clear how this is a notational aspect (a very convenient one). The components of a tensor are labeled with $p+q$ indices right? But if I present you with this list of numbers $$T_{i \,j\, k\, p\, l\, m\, n\,...}$$, can you guess to which kind of tensors these components belong to? No. So how can we encode this information? By distinguishing the indices! We put the ones involved in the covector expansions on the bottom and the ones related to vector expansions on top, but of course this is just convention. There's nothing fundamentally different about having indices on top and on the bottom, they just signal the information about the type of tensor, and it could easily be the other way around.

So that's the meaning of lower and upper indices: they tell us what type of tensor the components refer to. It turns out that choosing whether to put ones upstairs or downstairs fixes everything: vector components are upstairs, their basis downstairs, while for covectors its the other way around. And eventually we get tired of writing sums and the Einstein convention kicks in (and of course it only works because every sum in these tensor setting comes from expanding on a basis, so we'll always have the right pairing of upstairs and downstairs indices). Then,

What is the meaning of raising and lowering indices?

Well, it turns out finite-dimensional vector spaces are "boring" (and they're our friends because of that). You hopefully know that if two vector spaces have the same finite dimension, then they're isomorphic, which means, they are essentially the same, every member of one corresponds to a **unique ** member in the other. You can prove that the dual space of a finite dimensional vector space has the same dimension as that space and as such is isomorphic to it. Ok, but what does that mean physically? It means that every vector determines uniquely a covector, and that every covector determines uniquely a vector. For example, say you are talking about position in $3$ dimensions. $X$ is a vector right? But now construct these maps, called projections, \begin{align*} \pi_1(X)=X_1 \\ \pi_2(X)=X_2 \\ \pi_3(X)=X_3\end{align*} Then you can prove that for every vector position, the covector $\xi= X_i \pi^i$ corresponds uniquely to it. Physically this means that you can either define the position of something by a displacement vector or by the action of some map. Of course in the case of position this isn't very useful. But you'll notice that I wrote the components of $\xi$ as $X_i$ rather than $X^i$, because in the latter they occur as components of a $(1,0)$ type tensor. So you might say the isomorfism between a vector and a covector "lowers" the index, in the sense that the components after you "translate" the vector to a covector have now downstair indices.
So is it only a matter of switching the index? That, depends on the unique isomorphism you've chosen. It turns out $\mathbb{R}^3$ is a bad example because the isomorphism is rather simple. And so we come to your last question

Why does multiplying a tensor by a metric lower / raise index? What's the proof?

The proof is a bit technical, but the reason is really not; the metric gives the isomorphism! It tells you how to correspond the vectors to covectors. How? Say you want to convert the vector $v$. Then, consider this map $$g(v,\cdot): V \rightarrow \mathbb{R}$$ This certainly is a covector; it eats the vector $w$ to give the number $g(v,w)$. One can prove that this correspondence is unique (more than that, it is an isomorphism). Lets call this map $\beta$. Where does the whole businesses of multiplying by the metric lowering the index come from? From here: expand $\beta$ in a basis to give $\beta=\beta_i \epsilon^i$ and notice, what was $\beta$ in terms of the metric? It was $$g(v,w)=g_{ij}v^i\,w^j=(g_{ij}v^i)(w^j)=g_{ij}v^i \epsilon^j(w)$$ So now comparing these, what do you conclude about the components of $\beta$? $$\beta_i=g_{ij}v^j$$ Considering $\beta$ is in a one to one correspondence to $v$ then makes it reasonable to call its components, instead of $\beta_i$, $v_i$ such that you have $$v_i=g_{ij}v^j$$ that is, you "lowered the index". That's it.

I'm sorry if this explanation is not as simplistic as perhaps you would have hoped, but tensors are specific objects with specific theory; I'd say you can't understand them unless you at least understand the spirit and the general strokes of their formulation. So if you take away anything from this answer, take this:

  • The position of indices refers to what type of tensor the components pertain to, what kind of action it represents
  • The metric tensor gives a way of relating vectors and covectors
  • Raising and lowering indices arises when turning the components of one into the other, and the metric is involved because the metric is the thing that relates them

Remark: all these considerations can be applied to raising the index if you substitute the metric with the inverse metric, because if the metric gives the isomorphism to convert a vector into a covector, then the inverse of that isomorphism, that is, the inverse of the metric, tells you how to convert a covector into a vector. This also applies to more general tensors than vectors and covectors, but then it's more difficult to see why; nevertheless, the reason is the same, the metric gives an isomorphism between tensors of one kind and tensors of the other.

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All these questions are answered by differential geometry (See Geometrical Methods of Mathematical Physics, by Schutz for a great introduction).

What's the difference between lower and upper index of a tensor?

In differential geometry one has to throw away everything one knows about euclidean geometry and usual vector calculus and start to generalize the objects we knew, such that they also work in curved spacetime. When doing that the first step is usually vectors.

Vectors

We are used to defining a vector as the distance between two points. However, in differential geometry we might not have a well defined distance between two points. It is easier to "take the limit when the points goes to zero"and use derivatives as the basis for vectors. Note that a derivative also points in some direction, i.e. $\partial_x$ is as useful for a vector basis as $\hat{x}$. We then define

\begin{equation} \vec{V}=V^\mu \partial_\mu \end{equation}

Here $\vec{V}$ is the geometrical object, which is independent of coordinates, while $V^\mu$ is the $\mu$ coordinate in the basis of $\partial_\mu$. I'm using Einstein summation where

\begin{equation} V^\mu \partial_\mu =V^t \partial_t +V^x \partial_x +V^y \partial_y +V^z\partial_z \end{equation}

$V^\mu$ and $\partial_\mu$ transform in such a way that $\vec{V}$ is independent of the choice of coordinates.

Vectors are useful because they give us the notion of directional derivative. Indeed, vectors act on functions as

\begin{equation} \vec{V}(f)=V^\mu \partial_\mu f \end{equation}

So it gives us the rate of change of the scalar function $f$ along the direction $V^\mu$.

1-forms

Next, one might be interested in calculating lengths, surfaces, volumes, etc. The natural object is the differential $dx^\mu$, i.e., the object that goes inside an integral to tell you over which hypersurface you are integrating over. Note that, just as with vectors, we are using an infinitesimal object, differentials, as our basis (in the case of vectors it was derivatives). These objects are called p-forms, and a 1-form is

\begin{equation} \tilde{\omega}=\omega_\mu dx^\mu \end{equation}

Again, $\tilde{\omega}$ is coordinate independent, $\omega_\mu$ are the components of the 1-form in the basis of differentials $dx^\mu$. $\omega_\mu$ and $dx^\mu$ transforms in such a way that $\tilde{\omega}$ becomes independent of coordinates.

p-forms are useful to give us lengths,surfaces, volumes, etc. Indeed, they act on vectors

\begin{equation} \tilde{\omega}(\vec{V})=\omega_\mu V^\mu \end{equation}

where I used the relation $dx^\mu (\partial_\nu)=\delta^\mu_\nu$. There's a sense in which $\tilde{\omega}$ is a measure of length along the direction $\omega_\mu$.

What is the meaning of raising and lowering indices?

One might be tempted to ask "How do I measure the length of the vector $\vec{V}$? and answer "with a 1-form $\tilde{\omega}$" but which one?. Indeed, 1-forms measure length but there is no prefered 1-form to do so. What we need is a map that tells us which notion of length is "parallel" to which vector, sort of speak. This might seem obvious but it very much is not. This is equivalent to identifying column vectors with row vectors. Column vectors are vectors and row vectors are 1-forms, and the map between them is keeping all the components are the same. In special relativity, the map is flipping the sign of the time coordinate. In quantum meechanics, if you are familiar with that, we have kets $|\psi\rangle$ (the vectors) and bras $\langle \psi|$ (the 1-forms) and the map is doing thte complex conjugate.

In general, the object that tells you which 1-form goes with which vector to measure its length is the metric tensor $g_{\mu\nu}$ so that

\begin{equation} g_{\mu \nu}V^\mu V^\nu=V^2 \end{equation}

From this we can immediately read that the 1-form that we identify with the vector $\vec{V}$

\begin{equation} \tilde{V}=g_{\mu\nu}V^\nu dx^\mu \end{equation}

has component $V_\mu =g_{\mu\nu}V^\nu$. This is where lowering and raising indices comes from.

Which part of it is notation / convention and which part has actual mathematical and physical meaning?

It is always physical, but sometimes it doesn't look like it. For example, in the usual $\mathbb{R}^3$ space, column and row vectors have the same components, so using indices might seem pointless. Even in special relativity, lowering indices is just flipping the sign of the time component, which doesn't look like much. But it is always a physical transformation and the place of the index tells you how the object transforms and what it is supposed to do.

Why does multiplying a tensor by a metric lower / raise index? What's the proof?

There is no proof in the sense that the metric is built to do that. It has to satisfy certain properties of inner products. $V^2$ should be coordinate independent, it should be definite positive, etc. Also, the very definition of the metric

\begin{equation} ds^2=g_{\mu\nu}dx^\mu dx^\nu \end{equation}

is to give the "length" of $dx^\mu$.

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Upstairs indices label the components of a vector- $v^\alpha$ are the components of the vector $\vec{v}$. Downstairs indices label the components of a one-form- $\omega_\beta$ are the components of $\tilde{\omega}$. We define a new one-form $\tilde{v}$ (perhaps a poor choice in notation because we already have the vector $\vec{v}$) as $\boldsymbol{g}(\vec{v},\;)$. With this clever definition, the inner product of $\vec{v}$ with any arbitrary vector $\vec{u}$ is the same as the one-form $\tilde{v}$ acting on $\vec{u}$. $$\vec{v}\cdot\vec{u}=\tilde{v}(\vec{u})=g_{\alpha\beta}v^\alpha u^\beta = v_\beta u^\beta$$

So we’ve defined a new one-form $\tilde{v}$, with components $g_{\alpha\beta}v^\alpha=v_\beta$, that is not the vector $\vec{v}$, yet it acts almost the same way but in the guise of a one-form. This way of raising/lowering indices keeps the structure of the original object while allowing you to change between different types of tensors. In fancier lingo, the metric provides an isomorphism between the tangent and cotangent spaces.

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Lets try to make it very simple. If $$\mathbf{A} = A^{\mu} \mathbf{e}_{\mu}$$ is a vector, it is supposed to be an 'invariant' quantity in the sense that it doesn't matter what reference frame (basis) we express it in, it should always appear the same.

This is expressed by saying that if the basis $\mathbf{e}_{\mu}$ transforms under a matrix/transformation $M_{\mu}^{\ \ \nu}$ into $\mathbf{e}_{\mu}' = M_{\mu}^{\ \ \nu} \mathbf{e}_{\nu}$ then the $A^{\mu}$ should transform under the inverse, $(M^{-1})_{\mu}^{\ \ \nu}$, so that $\mathbf{A}$ remains invariant: $$\mathbf{A}' = A'^{\mu} \mathbf{e}_{\mu}' = (M^{-1})_{\rho}^{\ \ \mu} A^{\rho} M_{\mu}^{\ \ \sigma} \mathbf{e}_{\sigma} = A^{\rho} \delta_{\rho}^{\sigma} \mathbf{e}_{\sigma} = A^{\rho} \mathbf{e}_{\rho} = A^{\mu} \mathbf{e}_{\mu} = \mathbf{A}$$ Thinking of the basis as a frame of reference, and a transformation on the basis as a change of reference, we interpret the transformation on the basis as a covariant transformation, i.e. we co-vary with the basis/reference-frame, thus $\mathbf{e}_{\mu}$ is a covariant quantity, and things that transform contrary to this, i.e. under the inverse, like $A^{\mu}$, are contravariant quantities.

From this setup you see that $ \mathbf{A} \cdot \mathbf{e}_{\nu} = A^{\mu} \mathbf{e}_{\mu} \cdot \mathbf{e}_{\nu} = A^{\mu} g_{\mu \nu} := A_{\nu}$ is a covariant quantity, where I defined $g_{\mu \nu} = \mathbf{e}_{\mu} \cdot \mathbf{e}_{\nu}$, and assuming $g_{\mu \nu}$ is invertible with inverse $g^{\mu \nu}$ satisfying $g^{\mu \nu} g_{\nu \rho} = \delta^{\mu}_{\rho}$, we have $$\mathbf{A} = A^{\mu} \mathbf{e}_{\mu} = A^{\mu} \delta_{\mu}^{\nu} \mathbf{e}_{\nu} = A^{\mu} g_{\mu \rho} g^{\rho \nu} \mathbf{e}_{\nu} := A_{\rho} \mathbf{e}^{\rho} $$ leading to the notion of a dual basis $\mathbf{e}^{\rho} = g^{\rho \mu} \mathbf{e}_{\mu}$, where this satisfies $$\mathbf{e}^{\mu} \cdot \mathbf{e}_{\nu} = \delta^{\mu}_{\nu}$$ which lets me see immediately that $$A^{\mu} = \mathbf{e}^{\mu} \cdot \mathbf{A} = \mathbf{e}^{\mu} \cdot A^{\nu} \mathbf{e}_{\nu} \ \ , \ \ A_{\mu} = \mathbf{e}_{\mu} \cdot \mathbf{A} = \mathbf{e}_{\mu} \cdot A_{\nu} \mathbf{e}^{\nu} \ \ , $$ and I also have an expression for the contravariant components of $\mathbf{A}$ as $$A^{\mu} = \mathbf{e}^{\mu} \cdot \mathbf{A} = \mathbf{e}^{\mu} \cdot \mathbf{e}^{\nu} A_{\nu} = g^{\mu \nu} A_{\nu}.$$

In practice, the vectors $\mathbf{e}_{\mu}$ are defined by starting from the position vector $\mathbf{r} = x^{\mu} \hat{i}_{\mu}$, where $\hat{i}_{\mu}$ are the Cartesian unit basis vectors, putting the $x^{\mu}$ in curvilinear coordinates $u^{\mu}$ (e.g. polar/spherical coordinates etc...), and then setting $\mathbf{e}_{\mu} = \frac{\partial \mathbf{r}}{\partial u^{\mu}}$. Thus $$d \mathbf{r} = \mathbf{e}_{\mu} du^{\mu}$$ and so (in more familiar coordinates denoted $u^{\mu} = x^{\mu}$) the squared differential arc length is $$ds^2 = d \mathbf{r} \cdot d \mathbf{r} = \mathbf{e}_{\mu} \cdot \mathbf{e}_{\nu} dx^{\mu} dx^{\nu} = g_{\mu \nu} dx^{\mu} dx^{\nu}$$ For example, in special relativity we find $$ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 = (cdt,dx,dy,dz) \cdot (cdt, dx,dy,dz) = (cdt,dx,dy,dz) \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{bmatrix} \begin{bmatrix} cdt \\ dx \\ dy \\ dz \end{bmatrix} = (cdt,dx,dy,dz) \begin{bmatrix} cdt \\ -dx \\ -dy\\ -dz \end{bmatrix} = dx^{\mu} dx_{\mu} $$ so $g_{\mu \nu} = \mathrm{diag}(1,-1,-1,-1)$ and so if $$d \mathbf{r} = (cdt,dx,dy,dz) = dx^{\mu} \mathbf{e}_{\mu} = cdt(1,0,0,0) + dx (0,1,0,0) + .. + dz(0,0,0,1)$$ then $$d \mathbf{r} = (cdt,dx,dy,dz) = dx_{\mu} \mathbf{e}^{\mu} = cdt(1,0,0,0) - dx (0,-1,0,0) - .. - dz(0,0,0,-1)$$ thus $$dx_{\mu} = g_{\mu \nu} dx^{\nu} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{bmatrix} \begin{bmatrix} cdt \\ dx \\ dy \\ dz \end{bmatrix} = \begin{bmatrix} cdt \\ -dx \\ -dy\\ -dz \end{bmatrix} $$ where everything agrees.

Thus: the difference between an upper and a lower index is that the upper transforms under an inverse/contravariant basis transformationwhile a lower transforms under a direct/covariant basis transformation; raising and lowering indices just amounts to switching between expressing a vector as $\mathbf{A} = A^{\mu} \mathbf{e}_{\mu} = A_{\mu} \mathbf{e}^{\mu}$ i.e. working either with components transforming under the direct/covariant or inverse/contravariant transformation of the basis; the notation is basically just a convention that lets us talk about performing a direct transformation on the components (which means we are performing the inverse of the transformation that was applied to the basis/reference-frame); I gave a proof of why the metric raises the indices when I worked out $\mathbf{A} = A^{\mu} \mathbf{e}_{\mu} = A_{\mu} \mathbf{e}^{\mu}$ by inserting the $\delta$.

Let's quickly translate some of the other answers. Note that $A^{\mu} = \mathbf{A} \cdot \mathbf{e}^{\mu}$ is basically treating the vector $\mathbf{A}$ as though it were a `linear functional' of the basis $\mathbf{e}^{\mu}$, i.e. $A^{\mu} = \mathbf{A} \cdot \mathbf{e}^{\mu} := \mathbf{A}(\mathbf{e}^{\mu})$, i.e. a rank-1 tensor (a vector) is a linear functional of one vector, thus a rank $m$ tensor is a multi-linear functional $\mathbf{A}(\mathbf{e}^{\mu_1},...,\mathbf{e}^{\mu_m}) = A^{\mu_1 .. \mu_m}$. For example, the metric is a rank-2 tensor $g_{\mu \nu} = g(\mathbf{e}_{\mu},\mathbf{e}_{\nu}) = \cdot (\mathbf{e}_{\mu},\mathbf{e}_{\nu}) = \mathbf{e}_{\mu} \cdot \mathbf{e}_{\nu}$. Note that the usual (Euclidean space/Pythagorean theorem) $ds^2 = d \mathbf{r} \cdot d \mathbf{r} = (dx,dy,dz) \cdot (dx,dy,dz) = dx^2 + dy^2 + dz^2 = dx_{\mu} dx^{\mu}$ implies $dx_{\mu} = \delta_{\mu \nu} dx^{\nu} = dx^{\mu}$ i.e. there is no distinction between upper and lower indices in this special case due to the special form of the metric (compare to the special relativity case above). This is why you could write $w^a = g^{ab} w^b = g^{ab} w_b$ and get away with it in say fluids etc..., because you were working in standard 3D Euclidean space. Note I said that $\mathbf{e}_{\mu} = \partial_{\mu} \mathbf{r}$, if I throw away the position vector because I want to pretend to be very formal, I can say that $\mathbf{e}_{\mu} = \partial_{\mu}$, thus a vector would now read as $\mathbf{A} = A^{\mu} \partial_{\mu}$. Similarly, starting from $d f = dx^{\mu} \partial_{\mu} f = \mathbf{e}^{\mu} \partial_{\mu} f$ my dual basis is $\mathbf{e}^{\mu} = dx^{\mu}$, where $\mathbf{e}^{\mu} \cdot \mathbf{e}_{\nu} =\delta^{\mu}_{\nu}$ now reads as $dx^{\mu}(\partial_{\nu}) = \partial_{\mu} (dx^{\mu}) = \delta^{\mu}_{\nu}$. I could keep going, hopefully you've got the idea and can translate the rest of the other answers. See the Schaum's Tensor Calculus book for more details and examples, and these notes for a simple intro to some more of the theory.

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    $\begingroup$ Actually this helped a lot. But something that still confuses me. You said in Euclidean space the metric tensor is an identity, so I was not keeping track of raising and lowering of indicies. Does that mean in Euclidean space I can turn any vector into a covariant vector by multiplying by identity? Does that mean that I can turn velocity into a covariant vector by multiplying by Euclidean metric tensor (which is a Kronecker delta) and make a gradient of a scalar field to be contrvariant the same way? I am suspecting the answer is no, but I don’t understand why. $\endgroup$
    – Ilya Lapan
    Aug 17, 2023 at 20:07
  • $\begingroup$ @IlyaLapan Yes. For example, the gradient is defined as $\nabla f := \mathbf{e}^{\mu} \partial_{\mu} f = \mathbf{e}_{\mu} \partial^{\mu} f = \mathbf{e}_{\mu} g^{\mu \nu} \partial_{\mu} f$. In Euclidean space this means the components are $\partial_{\mu} f = \partial^{\mu} f$, i.e. $\frac{\partial f}{\partial x_{\mu}} = \frac{\partial f}{\partial x^{\mu}}$, which makes sense because I said $x^{\mu} = \delta^{\mu \nu} x_{\nu} = x_{\mu}$ so we can freely raise and lower indices in Euclidean space. $\endgroup$
    – bolbteppa
    Aug 18, 2023 at 3:36
  • $\begingroup$ In Euclidean space, when you take the dot product of two vectors, one way to think of what you are doing is that you are converting the second vector into a member of the dual space, and applying that resulting transform to the first vector. And you didn't have to change any of the coordinates to convert the vector to a transform. $\endgroup$ Aug 18, 2023 at 15:24
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Probably the easiest way to think of these things is to think of one-forms as differentials, and to think of vectors as directional derivatives. In some coordinate basis $x^{i}$, calculus classic tells us $\frac{d}{dx^{i}}\left(dx^{j}\right) = \delta_{i}{}^{j}$, so there is a natural pairing between vectors and one-forms baked into their definition that comes from this basic fact (the partial on the left is the vector basis, and it is pairing with the one-form basis, the differential on the right)

Now, let's look at flat space and kind of build an intuition. the metric gives us a pairing between the basis elements in the vector space, because in cartesian coordinates, we have the ${\vec v}\cdot {\vec w} = v_{1}w_{1} + v_{2}w_{2} + ...$, so the $v_{1}$ component of the vector pairs with the $w_{1}$ component of the other vector and so on. This pairing is governed by the fact that the metric is diagonal. The one-forms get a similar arrangement from the inverse metric.

So, if you want to ask "is there a one-form $dw$ that when acting on $\vec v$ gives me the same answer as ${\vec v}\cdot {\vec w}$, then, it should be clear that the answer is "yes! That one-form has components $w_{a} = g_{ab}w^{b}$!", and it should also be clear that if we want this to be an invertible process so that if we reverse this process, we have to mulitply by the inverse matrix on the left, so that we end up with $\vec w$ again.

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    $\begingroup$ I don't quite get why can't it be $w^{a} = g^{ab}w^{b}$ and what's the difference? $\endgroup$
    – Ilya Lapan
    Aug 17, 2023 at 16:06
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    $\begingroup$ In me 3 years of undergrad physics I would write all 3 indices at the top / bottom, not bothering with lower / higher indices (when doing E&M, fluids, etc). So $w^{a} = g^{ab}w^{b}$ looks ok to me. And $g^{ab}$ in flat space-time is just a matrix. So I don't see what changed now when raising and lowering them. Does it come down to covariant / contravariant vectors in the end? $\endgroup$
    – Ilya Lapan
    Aug 17, 2023 at 16:13
  • $\begingroup$ Suppose you have basis $\vec{e}_i$ and basis $\vec{f}_i$ with $\vec{e}_i = A^j_i \vec{f}_j$. Given a vector $\vec{v} = v^i \vec{e}_i$ with components $v^i$ in the $\vec{e}_i$ basis, we may write $\vec{v} = v^i A^j_i \vec{f}_j$. Thus $\vec{v}$ has components $v^j A^i_j$ in the $\vec{f}_i$ basis. The change of basis causes the change $v^i \to v^j A^i_j$ of components. Take dual basis $\sigma^i$ to $\vec{e}_i$ (i.e. $\sigma^i (\vec{e}_j) = \delta^i_j$) and dual basis $\eta^i$ to $\vec{f}_i$. What happens to the components of a general 1-form $\lambda$? (Continued) $\endgroup$ Aug 18, 2023 at 3:02
  • $\begingroup$ Suppose $\lambda = \lambda_i \sigma^i = \lambda_i ' \eta^i$. Clearly $\lambda(\vec{e}_i) = \lambda_i$. On the other hand $\lambda(\vec{e}_i) = \lambda(A^j_i \vec{f}_j) = A^j_i \lambda(\vec{f}_j) = A^j_i \lambda_k ' \eta^k(\vec{f}_j) = A^j_i \lambda_k' \delta^k_j = A^k_i \lambda_k ' $. That is, $\lambda_i = A^k_i \lambda_k '$, or $\lambda_i ' = [A^{-1}]^j_i \lambda_j$. The change of coordinates causes the change $\lambda_i \to [A^{-1}]^j_i \lambda_j$ of components. $\endgroup$ Aug 18, 2023 at 3:31
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    $\begingroup$ Our convention is that upper indices refer to the components of a vector and lower indices refer to the components of a 1-form. Could we have done it the other way around? Absolutely. Could we have said $v^i$ are components of a vector and used a hat like this $\hat{\lambda}^i$ for components of a 1-form without changing the position of the index? Absolutely. The point is that we need some way to keep things straight because we have two things which transform in fundamentally different ways. Upper and lower indices are just the way all mathematicians and physicists have decided to keep track. $\endgroup$ Aug 18, 2023 at 3:40
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I think that a particular example can be useful for an intuitive grasp of the concepts. The picture is after the text.

If two axis X1 and X2 in a plane are not orthogonal, but have an angle $\phi$, there are 2 ways to define the components of a vector, represented by an arrow joining the origin O to a point P:

X1-component can be defined as the length resulting from parallels to X2 axis from P until cross X1 axis ($X^1$), and X2-component the length resulting from parallels to X1 axis from P until cross X2 axis ($X^2$). The picture is a parallelogram, and it can be understood as the usual vectorial sum of a linear combination of basis vectors of each axis.

The components can alternatively be defined as the lengths resulting from projecting OP in each axis, tracing a perpendicular ($X_1$) and ($X_2$).

The first mode is called contravariant components, and the second mode covariant components.

The square of the length OP, invariant by axis rotations, is $$OP^2 = (X^1)^2 + (X^2)^2 + 2X^1X^2 cos(\phi)$$ But it can be shown geometrically that $$OP^2 = X^1X_1 + X^2X_2$$

That means: applying the dot product rule of a contravariant by a covariant vector results on an (invariant) scalar.

So, a contravariant vector is a kind of machine that takes as input a covariant vector and produces a scalar. The inverse for a covariant vector.

A rank 2 covariant tensor takes as input 2 contravariant vectors and produces a scalar. An example is the metric tensor, which can be checked in this example. The result is exactly the invariant $OP^2$. By the way, when of the first input, the metric tensor produces a covariant vector. The second input is the already mentioned covariant vector taking a contravariant one as input, and generating the scalar $OP^2$

enter image description here

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