# How to visualize the gradient as a one-form?

I am reading Sean Carrol's book on General Relativity, and I just finished reading the proof that the gradient is a covariant vector or a one-form, but I am having a difficult time visualizing this. I usually visualize gradients as vector fields while I visualize one-forms with level sets. How to visualize the gradient as a one-form?

• Why do you visualize 1-forms as level sets? – DanielSank Jan 14 '15 at 6:13
• I visualize 1-forms as level sets because someone advised me to when I asked how to visualize 1-forms on the mathematics stack exchange. – user70720 Jan 14 '15 at 6:25
• Schutz's A first course in General Relativity explained this nicely. – jinawee Jan 14 '15 at 11:25
• You're free to visualize n-forms as n-vector fields (1-forms as vector fields). This may be easier than imagining level sets. en.m.wikipedia.org/wiki/Multivector – Luke Burns Jan 10 at 0:27

If you're going to take that path, then maybe you should be thinking more of a level set density, i.e. how closely spaced the level sets in question are. Sean Carrol's book is not wonted to me: if you can get a copy of Misner Thorne and Wheeler, the first few pages do a good job of this idea with their quaint "bong" machine that sounds a "bong" bell each time a vector pierces a level set. If you can't get this readily, then the early part of Kip Thorne's lectures here is also good

Anyhow, suppose we are given a scalar field $\phi(\vec{x})$ and the tangent space $T_x\mathcal{M}$ to $x\in\mathcal{M}$ in some manifold $\mathcal{M}$, and we imagine riding along a vector $X\in T_x\mathcal{M}$ in the tangent space: how often would we pierce level sets of $\phi$: in MTW's quaint and unforgettable words (and wonderful sketches), how often would our bell sound as we rode along the vector? It would be $\nabla\phi\,\cdot\,X$ (i.e. the directional derivative). Thus $\nabla\phi$ is a dual vector to the vector space of tangent vectors. It is a linear functional $T_x\mathcal{M}\to\mathbb{R}$ on the tangent space: it takes a vector $X\in T_x\mathcal{M}$ as its input and spits out the directional derivative $\nabla\phi\,\cdot\,X$.

• Thank you very much. This helps a lot. The reason I use sean carrol's book is because it is free, though he doesn't use tools like the Bong machine to facilitate visualization of concepts. – user70720 Jan 14 '15 at 6:32
• I think there is a big typo in this answer. The first paragraph just sort of... dies. – DanielSank Jan 14 '15 at 6:54
• @DanielSank Thanks. I meant to cite some of Kip Thorne's lectures that the OP can get on the web, and then forgot to do this! – WetSavannaAnimal Jan 14 '15 at 7:46
• @Andy I added a link to Kip Thorne's lectures, which are somewhat like MTW in their style. – WetSavannaAnimal Jan 14 '15 at 7:47
• @DanielSank BTW great photo for your user icon. Did you take this yourself: what species is this beastie - it looks a little like a Hapalochlaena species. – WetSavannaAnimal Jan 14 '15 at 10:44

Given a smooth manifold $$M$$, let $$(U,\phi)$$ be any chart on it, and let $$X$$ be a vector field. Then $$X$$ can be interpreted as a derivation if you think of every vector $$X_p$$ of the field as being the velocity vector of some path through $$p$$ with tangent vector $$X_p$$ itself. Now with the help of the coordinates given by the chosen chart, the vector field on $$U$$ can be written as $$X = \sum_{k=1}^n X^k\frac{\partial}{\partial x^k},$$ where the $$X^k$$ can be seen as the contravariant components of $$X$$ w.r.t. the chosen chart. The derivatives $$\frac{\partial}{\partial x^k}$$ are, at any point of $$U$$, a base of the tangent space $$T_pM$$. Geometrically, they are the tangent vectors to the curves on the manifold created by the chart under consideration (i.e. the curves $$\gamma_i(t) = \phi^{-1}(\phi(p) + e_it)$$, $$e_i$$ being the canonical basis vector of $$\mathbb R^n$$) at $$p$$. This notation comes from the fact that, if $$f$$ is any smooth function from $$M$$ to $$\mathbb R$$, $$X$$ operates on $$f$$ as a directional derivative $$X(f)$$, and in components on $$U$$ this is precisely $$X(f) = \sum_{k=1}^n X^k\frac{\partial f}{\partial x^k},$$ Now this expression can be dualised, in the sense that you can think that it is $$f$$ that is acting on the vector field through its transpose $$f^*$$ that is such that $$f^*(X) = X(f)$$ If you denote by $$\text d x^k$$ the dual basis of $$\frac{\partial}{\partial x^k}$$ on the cotangent bundle $$T^*M$$, it follows by a direct computation that $$f^* = \sum_{k=1}^n\frac{\partial f}{\partial x^k}\text d x^k,$$ which suggests the more evocative notation $$\text df = f^*$$. The components of $$\text d f$$ which is clearly a 1-form, are precisely those of the gradient, i.e. the partial derivatives of $$f$$.

As a further remark, observe that the gradient is usually defined as a vector field through the musical isomorphisms (in case then that you have a (pseudo)Riemannian manifold) induced by a metric tensor. More precisely, the gradient vector field of a smooth function $$f$$ is defined by $$g(\nabla f, Y) = \text d f(Y) = Y(f)$$ for any vector field $$Y$$ on $$M$$.

• "musical" isomorphism? Your own phrase? – DanielSank Jan 14 '15 at 17:46
• – Phoenix87 Jan 14 '15 at 23:45

In the context of general relativity it is also notable that the manifold is equipped with a metric tensor. This tensor provides a unique way (an isomorphism, s.a. answer by Phoenix87) to map covectors to vectors. The components are computed by index raising/lowering, e.g. for a given covector $w$ with components $w_i$ the corresponding vector reads $v^i=g^{ij}w_j$, where $g^{ij}$ are the components of the metric tensor and I use the summation convention over repeated indices.

Now the question is how this really looks like in the given case. The natural derivative of a function $f$ on a manifold $\mathcal{M}$ is $df$, with components $df_i = \partial_i f$, since this is possible without metric. As already mentioned, $df$ is visualized by level sets of $f$. Now the corresponding contravariant vector is computed by $X^i := g^{ij}\partial_i f\equiv (\nabla f)^i$, which we identify with the gradient of $f$. This vector is the vector that is everywhere perpendicular to the level sets of $f$. Note that the notion "perpendicular" can only be defined with a metric.

Many textbooks in physics also visualize the basis vector $dx^i$ with its gradient vector, i.e. the vector that is everywhere perpendicular to the levels sets of $x^i$. This is strictly speaking only possible when there is a metric present, but it simplifies the comparsion to $\frac{\partial}{\partial x^i}$ which is in general not the same direction.