I have been trying to convince myself that it is consistent to replace basis vectors $\hat{e}_\mu$ with partial derivatives $\partial_\mu$. After some thought, I came to the conclusion that the basis vectors $\hat{e}_\mu$ were ultimately just symbols which represent what we think of as arrows, so it is not a problem to use a different symbol. The only requirement is that one can manipulate the $\partial_\mu$ in the same way as the $\hat{e}_\mu$.

However, raising/lowering indices seems to create an inconsistency. In switching our representation of the basis vectors, we make the substitutions:

$$\hat{e}_\mu \rightarrow \partial_\mu$$

$$\hat{e}^\mu \rightarrow dx^\mu$$

However, while we previously could write $\hat{e}^\mu=g^{\mu \nu}\hat{e}_\nu$, we fail to be able to write the same relationship in the new representation:

$$dx^\mu \neq \partial^\mu =g^{\mu \nu} \partial_{\nu}$$

My questions are:

  • Have I done something invalid here?
  • If not, is it just an unwritten rule that one should never try to raise an index of a basis vector?
  • What is the motivation to write basis vectors as partial derivatives or differentials (for the tangent or cotangent space) as opposed to just writing some other symbol? Do we actually need the properties of a derivative or differential in our basis vectors? I am aware that the $\partial_\mu$ resemble the expression $\frac{\partial\vec{r}}{dx^\mu}$ which is a natural choice for the basis vectors $\hat{e}_\mu$, but the differentials seem to come out of nowhere.
  • $\begingroup$ Could you explain the justification behind "we previously could write $\hat{e}^\mu = g^{\mu \nu} \hat{e}_\nu". This equation seems manifestly wrong, because you can't write a covector as a linear combination of vectors. $\endgroup$ May 19, 2017 at 14:27
  • $\begingroup$ @AccidentalFourierTransform I think it depends on what your starting point is. If you start with a smooth manifold with a coordinate chart having coordinates $x^\mu$, and you define vectors as differential operators on functions, then the vector $\partial_\mu$ is a derivative operator, and is in fact the derivative operator given by partial differentiation with respect to $x^\mu$: $\frac{\partial}{\partial x^\mu}$. $\endgroup$ May 19, 2017 at 14:31
  • $\begingroup$ @AccidentalFourierTransform I have seen examples where the derivatives are used as operators, namely in Nakahara, Geometry, Topology, & Physics. Here is a screenshot, but it's hard to get it out of context: imgur.com/a/hhopb A friend of mine also notes they are used to define Lie Brackets. I haven't studied Lie Brackets, so I don't have much to say on that. $\endgroup$ May 19, 2017 at 14:51
  • $\begingroup$ @doublefelix Don't worry about it being hard to get out of context; we have all seen those examples. He(AccidentalFourierTransform) already knows what you are talking about. $\partial_\mu$ exactly represents a differential operator in exactly the sense you linked. $\endgroup$ May 19, 2017 at 14:54
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Justification for $\hat{e}^\mu = g^{\mu \nu} \hat{e}_\nu$: In general one can always raise or lower an index to find the covariant or contravariant counterpart to a lorentz index. For example, we have for any vector that $x = x^\mu \hat{e}_\mu = x_\mu \hat{e}^\mu$. Basis vectors are no exception, and here it seems relevant because I am asking about the basis for the cotangent space $dx^\mu$. $\endgroup$ May 19, 2017 at 14:55

4 Answers 4


Raising and lowering indices in a vector is not a valid operation. Basis vectors are no exception. While $x_\mu=g_{\mu\nu}x^\nu$ is a valid operation, $\hat e^\mu=g^{\mu\nu}\hat e_\nu$ is not. The reason is that in the first case you are dealing with the components of a vector, and in the second case you are dealing with a vector itself.

Let me elaborate. Given a vector $\hat X$ $$ \hat X=x^\mu\hat e_\mu $$ you can lower the index $\mu$ in $x^\mu$ through $$ x_\mu\equiv g_{\mu\nu}x^\nu $$

That is: raising and lowering indices is an operation that is defined for the components of a vector (or covector).

The index $\mu$ in $\hat e_\mu$ is not a vector index; it just labels the different basis vectors. You cannot raise/lower this index, because $\hat e_\mu$ does not denote the components of any vector. The operation $$ \phantom{\color{red}{\text{NO!}}}\qquad\hat e^\mu\equiv g^{\mu\nu}\hat e_\nu\qquad\color{red}{\text{NO!}} $$ is a meaningless operation.

The same thing can be said about covectors. Given a covector $\tilde X$ $$ \tilde X=x_\mu\tilde e^\mu $$ you can raise the index in $x_\mu$. But you cannot lower the index in $\tilde e^\mu$, because that index does not denote the components of a covector; it just labels the different basis covectors.

Most importantly, while $\hat e_\mu$ is a basis of the space of vectors, and $\tilde e^\mu$ is a basis for the space of covectors, these objects are not related through $$ \phantom{\color{red}{\text{NO!}}}\qquad\hat e^\mu= g^{\mu\nu}\tilde e_\nu\qquad\color{red}{\text{NO!}} $$ or any similar relation.

In short: you can raise/lower indices when those indices denote the components of an object - either a vector or a covector - but you cannot raise/lower the indices of the bases of vectors/covectors, because those indices do not denote the components of anything. They are just labels.

However, see Musical isomorphism.

I hope that at this point, you are still with me. Given an arbitrary vector $\hat v$ (like $\hat X$ or $\hat e_\mu$), and a certain function $f$, we can define the action of $\hat v$ on $f$ as follows: we define $$ \hat e_\mu[f]\equiv \frac{\partial f}{\partial x^\mu}\in\mathbb R $$ and we extend this through linearity: if $\hat v=v^\mu \hat e_\mu$, then $$ \hat v[f]=v^\mu\frac{\partial f}{\partial x^\mu}\in\mathbb R $$

I'm not going to discuss why this new operation is useful. But let me stress that this operation is something new, something that you might have never seen before: now vectors can act on functions! In any case, useful or not, this new operation motivates us to consider the following convenient notation: we will write $\hat \partial_\mu$ instead of $\hat e_\mu$: $$ \hat \partial_\mu\equiv \hat e_\mu $$

With this, our equation from before now becomes $$ \hat\partial_\mu[f]=\frac{\partial f}{\partial x^\mu} $$

Note that we are using the same symbol, $\partial$, with two different meanings: on the one hand, it denotes a basis vector, and on the other hand, it denotes a partial derivative. The usual thing we do is to drop the distinction: we just write $\partial_\mu$ for both, and let context decide what the symbol means.

In the same vein, we usually use the symbol $\mathrm dx^\mu\equiv\tilde e^\mu$. That is, we denote the basis of covectors by the symbol $\mathrm dx^\mu$. It's just notation.

Let us now move on to the gradient. We define the covector $\mathrm d f$ as the covector that has $\frac{\partial f}{\partial x^\mu}$ as components: $$ \mathrm d f=\frac{\partial f}{\partial x^\mu}\tilde e^\mu $$ or, using our new notation, $$ \mathrm d f=\partial_\mu f\,\mathrm dx^\mu $$

You can raise and lower the $\mu$ index in $\frac{\partial f}{\partial x^\mu}$, because this index denotes the components of a covector. In this sense, you could say that you can raise/lower the $\mu$ index in $\partial_\mu$, whenever this symbol denotes a derivative. But you cannot raise/lower the $\mu$ index in $\hat \partial_\mu$, whenever this symbol denotes a basis vector (for the same reason you cannot raise/lower the $\mu$ index in $\hat e_\mu$).

In short: the objects $\partial_\mu$ and $\mathrm dx^\mu$ replace the old notation $\hat e_\mu$ and $\tilde e^\mu$, but they denote the exact same object: they are a basis for the space of vectors and covectors. This means that you cannot raise/lower their indices. On the other hand, the object $\partial_\mu f$ denotes the components of the covector $\mathrm df$, and as such, you can raise/lower its index.

  • $\begingroup$ Then, is it not true that a vector v = $v=v^\mu \hat{e}_\mu=v=v_\mu \hat{e}^\mu$? I know that for all concrete examples I have done this has been the case. And if this equation holds, then: $v=v_\mu \hat{e}^\mu =v^\mu \hat{e}_\mu=g^{\alpha \mu}v_\alpha \hat{e}_\mu=v_\mu(g^{\mu \alpha} \hat{e}_\alpha$) so it seems that any basis of the cotangent space $\hat{e}^\mu$ is equal to $g^{\mu \nu} \hat{e}_\nu$ $\endgroup$ May 19, 2017 at 16:47
  • $\begingroup$ The OP might have learned about covectors in the context of nonorthornormal bases in Euclidean space. In this case there's always a natural isomorphism between vectors and covectors and I think their formula is correct. $\endgroup$
    – knzhou
    May 19, 2017 at 16:56
  • $\begingroup$ @knzhou yes, that's the context I learned them in. So this no longer works outside of Euclidean space? I was under the impression that we always deal with vectors on tangent planes exactly for the reason that we can treat them in the way we did on flat space (at least until you try to relate vectors at different points, then you need the christoffel symbols etc) $\endgroup$ May 19, 2017 at 17:28
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    $\begingroup$ @doublefelix note that $v^\mu\hat e_\mu=v_\mu \hat e^\mu$ cannot be correct in general, because the l.h.s. is a vector, while the r.h.s. is a covector: those objects live in different spaces, so you cannot equate them. Of course, as knzhou mentions, in Euclidean space there is a natural way to identify vectors and covectors, so only in that case does that equatiion make sense (technically, it is not correct as written, but it can be fixed). $\endgroup$ May 19, 2017 at 17:34
  • $\begingroup$ In that case, I suppose the source of the problem is that I don't understand the fundamental difference between covectors and vectors. Having come from physics, I thought it was just their opposite transformation properties. I am aware of the mathematical formalism in which you see covectors as linear functionals of vectors, but thus far, this seemed just like another way of thinking about the same notion. $\endgroup$ May 19, 2017 at 18:02

To understand what happens when we raise or lower indices, we have to see what actually are the objects we are operating on.

TL;DR - You raise (lower) components of vectors (dual vectors), not their basis.

To see why derivatives are used as a basis, we use the following motivation: Imagine a curve somewhere in $\mathbf{R}^3$. The curve will have a tangent vector over each of its points. If we denote the curve as $r(\lambda)$ (a function) with $\lambda$ the curve parameter, the tangent vector will be $$\mathbf{t} = \frac{d\mathbf{r}(\lambda)}{d\lambda} = \sum\frac{dx^i}{d\lambda}\hat{x}^i$$ So now we know the "rate and direction of change" of the curve, at a point $r(\lambda)$. We have gained a vector, and we would like to use this vector to describe other things happening over that manifold.

The next question we ask, is what is the rate of change of some other object in the direction of that first vector. Well, we are still in $\mathbf{R}^3$, so we know how to find those "rates of change" - the nabla operator $\nabla = \frac{\partial}{\partial x} \hat{x} + \frac{\partial}{\partial y} \hat{y} + \frac{\partial}{\partial z} \hat{z}$. To find the rate of change in the direction of the previously gained vector, we project the $\nabla$ on $\mathbf{t}$

$$\mathbf{t}\nabla = \sum \frac{dx^i}{d\lambda}\frac{\partial}{\partial x^i}$$ Here we see that by using tangent vectors, we can probe and gain information about rates of change of objects in certain directions.

Now, if we use this motivation, that vectors, when acting on some objects, give us information about some rate of change, we can construct a basis for tangent vectors over $\mathbf{R}^3$ which is $\{ \frac{\partial}{\partial x} , \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \}$

We can show that such a basis can be constructed over each point of a general manifold, short-handed written $\{\partial_\mu\equiv\frac{\partial}{\partial x^\mu}\}$.

Vectors were constructed as objects acting on functions, and there are also objects that act on vectors and send them to real numbers, called dual vectors. Again, by drawing motivation from $\mathbf{R}^3$, we can construct a basis for these dual vectors, and we denote this basis as $\{ dx^\mu\}$, where this basis is defined by how it acts on the vector basis: $$dx^\mu(\partial_\nu) = \delta^\mu_\nu$$

Now comes one of the key points where the mistake happened - we have defined a basis for vectors, and a vector is an object such as $u = u^\mu\partial_\mu$. For instance, we can construct a vector $v=1\cdot\partial_1$. So, here number 1 is a vector component, while $\partial_1$ is the component of a basis. As an example, a dual vector would be written as $\omega = \omega_\nu dx^\nu$.

Vectors and dual vectors have a special relationship, a dual vector $\omega$ acts on a vector $v$ and sends it to a real number. We can express this using their bases. $$\omega(v) = \omega_\mu dx^\mu v^\nu\partial_\nu=\omega_\mu v^\nu dx^\mu\partial_\nu = \omega_\mu v^\nu\delta^\mu_\nu = \omega_\mu v^\mu$$

Now we come to the metric tensor. A tensor is such an object which acts on a certain number of vectors and dual vectors, depending on the tensor type. A metric tensor is a tensor which acts on two vectors.

We can write down the metric tensor using the previously defined bases as: $$g = g_{\mu\nu}dx^\mu \otimes dx^\nu$$ So the action of a metric tensor, is, it takes two vectors as an input and outputs a real number. Written fully in a basis this is: $$g(u, v) = g_{\mu\nu}dx^\mu(u^\alpha \partial_\alpha) \otimes dx^\nu (v^\beta \partial_\beta)$$ $$g(u, v) = g_{\mu\nu}u^\alpha v^\beta dx^\mu (\partial_\alpha) dx^\nu(\partial_\beta)$$ $$g(u, v) = g_{\mu\nu}u^\alpha v^\beta \delta^\mu_\alpha \delta^\nu_\beta = g_{\mu\nu}u^\mu v^\nu$$

This operation has a short-hand notation $$g_{\mu\nu}u^\mu v^\nu = u_\nu v^\nu$$ and only here is where the lowering and raising happens. This can also be formally well defined by saying that the metric induces a natural isomorphism between vectors and dual vectors.

So, when you lower indices, you must only act on components of vectors, not on the basis, likewise, when you raise indices you must only act on components of dual vectors, not their basis.

For a good reference, I recommend "An introduction to manifolds" by Loring W. Tu


I think your source of confusion is conflating the use of enumeration indices for basis vectors, and the use of vector indices for the components of a vector. These two types of indices need to be treated differently. First I will say what I mean by the two types of indices, then I will say how they need to be treated differently.

The first type of index is an enumeration index for the basis. So lets suppose we have $n$ dimensional vector space, and lets choose a basis. The basis vectors can be written as $$\hat{e}_\mu,\quad \mu = 1,2,3,\cdots,n.$$ In this case, the index $\mu$ is an enumeration index just used to list the basis vectors.

Now a vector $v$ can be written using coordinates with respect to this basis. In this case we would write $v=v^\mu \hat{e}_\mu$. In this case, the $\mu$ in $v^\mu$ is a vector index. The difference is that for each value of $\mu$, $v^\mu$ is just a number where $\hat{e}_\mu$ had been a vector. Additionally, if we change bases to new basis $\hat{\tilde{e}}_\mu$, related to the orginal basis $\hat{e}_\mu$ by $$\hat{\tilde{e}}_\nu = R_\nu{}^\mu \hat{e}_\mu, $$ then the coordinates $\tilde{v}^\nu$ of $v$ with respect to the new basis $\hat{\tilde{e}}_\nu$ are related to the old coordinates $v^\mu$ with respect to the old basis $\hat{e}_\mu$ by $$\tilde{v}^\nu = R^{-1}{}^\nu{}_\mu v^\mu.$$

I think by now I have explained how these two types of indices must be treated differently. One enumerates a set of vectors, the other enumerates a set of real number coordinates that transform under coordinate transformations.

Now lets suppose we have an inner product with coordinates $g_{\mu\nu}$. Then for any basis $\hat{e}_\mu$, you can obtain a dual basis $\hat{e}^\mu$ for the covector space, satisfying $$\hat{e}^\mu(\hat{e}_\nu) = \delta^\mu{}_\nu.$$

Now given any vector $v$, you can associate with a dual vector $v'$, where this dual vector $v'$ acts on vectors $w$ by taking the inner product with it $\langle v, w \rangle$. To get the coordinates of this dual product $v'$, we can write in in the form $v' = v'_\mu \hat{e}^\mu$. We find that $$ v^\mu g_{\mu \nu} = \langle v, \hat{e}_\nu \rangle = \langle v'_\mu \hat{e}^\mu, \hat{e}_\nu \rangle = v'_\mu\langle \hat{e}^\mu, \hat{e}_\nu \rangle = v'_\mu\delta^\mu{}_\nu = v'_\nu. $$ Therefore we find that if $v^\mu$ are the coordinates of a vector with respect to some basis, then the coordinates $v'_\mu$ of the dual vector $v'$ with respect to the dual basis is given by $v'_\nu = v^\mu g_{\mu \nu}$. In this sense, you can use the metric to raise the indices on coordinates. This is made possible because each coordinate is a real number, and when you take linear combinations of real numbers, you get another real number.

On the other hand, you cannot say $\hat{e}_\nu = \hat{e}^\mu g_{\mu \nu}$, because the right hand side is a vector, and the left hand side is a linear combination of covectors, which gives you another covector, but covectors and vectors are different kinds of objects, so they can't be equal.

I think this should answer your first two questions. I don't really know the answer to the third question other than to say that the easiest way of defining the tangent space is in terms of derivative operators and the partial derivatives with respect to coordinates make a natural basis.


Everything I have to say here pertains to coordinate bases with an associated metric.

I started to say that this is an incorrect statement:

$$\partial^\mu = g^{\mu \nu} \partial_{\nu},$$

but then realized it does have a valid interpretation. Just not one that typically occurs in the context of differential forms. See the end of this post.

This is correct:

$$dx^\mu =g^{\mu \nu} \partial_{\nu}.$$

The substitutions suggested in the original question are, in fact valid (without the hats):

$$e_\mu \rightarrow \partial_\mu,$$

$$e^\mu \rightarrow dx^\mu.$$

They may even be taken as definitions.

I only use hats to indicate orthonormal bases. At a minimum, the hats will indicate unit basis vectors. In curvilinear coordinates such a basis will not, in general be a coordinate basis. A distinction worth understanding.

Raising and lowering indices on basis vectors and 1-forms is legitimate. It's just a matter of associating the contraction of the metric with the basis vector or 1-form, rather than with the vector components:


etc. So $$v_{j}=v^{i}g_{ij},$$ and $$\mathfrak{e}_{i}=g_{ij}\mathfrak{e}^{j}.$$

If you don't like to call the contravariant basis vectors, "vectors", then call them "basis 1-forms".

From there, it's just a matter of applying the established definitions. If you lower the index on a basis 1-form (AKA contravariant basis vector) it becomes a (covariant) basis vector, and is therefore identified with a directional derivative. The directional derivative along a (covariant) basis vector is a partial derivative.

In the language of differential forms, $dx^{i}$ is the (location dependent) projection mapping of vectors onto the coordinate direction indicated by $i$:

$$\left\langle dx^{i},\mathfrak{v}\right\rangle =\mathfrak{e}^{i}\cdot\mathfrak{v}=v^{b}\mathfrak{e}_{b}\cdot\mathfrak{e}^{i}=v^{i}.$$

So a basis vector (1-form) with a raised index is to be interpreted as such a projection mapping. In other words,



$$\mathfrak{e}^{j}\equiv dx^{j},$$


On an orthonormal basis, we sometimes write $$\partial^{i}\equiv\delta^{ij}\partial_{j}=\partial_{i}.$$

At first I thought that $\partial^\mu = g^{\mu \nu} \partial_{\nu}$ would amount to mathematical gibberish. But I then realized that it does have a valid interpretation. Every linearly independent spanning basis at a point in our manifold has a corresponding dual basis, which the differential form school calls a 1-form basis. But in the dual vector basis school, this is just another coordinate basis spanning the tangent space, and there is an associated coordinate system. $\partial^\mu = g^{\mu \nu} \partial_{\nu}$ is the partial derivative operator along the contravariant basis vectors.

Regardless of whether the differential forms school admits that contravariant basis vectors and dual basis 1-forms are identical, both exist. There are, therefore, contravariant basis vectors which can be mapped uniquely to directional derivatives.

Now, the question becomes, can the projection mappings $dx^i$ be shown to mean the same thing as $\partial^i$? I doubt it.

But we could just as well invert the relationship between contravariant and covariant vectors, so that what we originally call "covariant basis vectors" become our new basis 1-forms and vice versa.

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    $\begingroup$ It would be helpful (and appropriate) if the person who down-voted my answer would explain that action. I have considerable experience in this area, and believe my answer is correct and valuable. If I am mistaken, then it is my wish to repair my understanding. $\endgroup$ Aug 10, 2018 at 2:45

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