5
$\begingroup$

Finding the Gradient of a Scalar Field

I understand that you can find the gradient of a scalar field, in an arbitrary number of dimensions like so :

$$grad(f) = \vec{\nabla}f = \left<\frac{\partial f}{\partial x_{1}}, \frac{\partial f}{\partial x_{1}},...,\frac{\partial f}{\partial x_{n}}\right> = \begin{bmatrix} \frac{\partial f}{\partial x_{1}} \\ \frac{\partial f}{\partial x_{2}} \\ ... \\ \frac{\partial f}{\partial x_{n}} \end{bmatrix}$$

where $f$ is a scalar function, $f: \mathbb{R^n} \to \mathbb{R}$. And as you can see this generalizes consistently to higher-dimensional scalar fields. All of this is straight out of Multivariable Calculus.

Finding the Gradient of a Vector Field

Furthermore finding the gradient of a Vector Field, is given by a Tensor i.e. given $f$ to be a vector function, $f : \mathbb{R^m} \to \mathbb{R^n}$ :

$$grad(\vec{f}) = \displaystyle \nabla \vec{f} = T = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & ... & \frac{\partial f_1}{\partial x_m} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & ... & \frac{\partial f_2}{\partial x_m} \\ ... & ... & ... & ... \\ \frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2} & ... & \frac{\partial f_n}{\partial x_m} \\ \end{bmatrix}$$

with $T$ denoting the tensor (a $n\ $x$\ n$ matrix of partial derivatives of $\vec{f}$'s scalar components, i.e. rank-$0$ tensor components, correct me if what I said in these brackets is wrong) which tells us how the vector field changes in any direction.


How to find the Gradient of a Tensor Field?

But how do you find the gradient of a Tensor Field? I understand that to answer this question we may need to generalize the concept of a tensor field a bit further.

If I understand correctly, a scalar is a tensor of rank-$0$, a vector is a tensor of rank-$1$. Is it then fine to generalize scalar fields as tensor fields of rank-$0$, and vector fields as tensor fields of rank-$1$? If so then it means we've been finding the gradient of tensor fields (albeit of rank-0 being scalar fields) in our Multivariable courses all along, we just didn't know it.

By extending the logic behind the leap between taking the gradient of a Scalar Field, to taking the gradient of a Vector Field, is it then correct to say that :

The gradient of a Tensor field of rank-$n$ is a Tensor field of rank-($n+1$) ?

$\endgroup$
  • $\begingroup$ I think what you're looking for is called the covariant derivative $\endgroup$ – Sir Jective Apr 22 '16 at 21:51
  • $\begingroup$ I'm not entirely sure if the matrix I included to denote the Tensor $T$ is fully correct, or if what I said below it is fully correct either. Could someone please verify that the matrix I've included and what I've said below it is correct, I wouldn't want other users to get confused by an error on my part. $\endgroup$ – Perturbative Apr 22 '16 at 21:52
  • $\begingroup$ btw, i think you mean to say vector fields as tensor fields of rank-$1$. $\endgroup$ – Sir Jective Apr 22 '16 at 22:04
  • $\begingroup$ @silvascientist. Yep, you're correct, fixed the error now. Have I made any other errors in my representation of the Tensor $T$ as a matrix? $\endgroup$ – Perturbative Apr 22 '16 at 22:19
3
$\begingroup$

A tensor is just a multilinear, scalar-valued function. If I write $V \multimap W$, for the collection of linear functions from a vector space $V$ to a vector space $W$, then, over the reals, a rank-$n$ tensor is just $V^{\otimes n}\multimap\mathbb R$ where $V^{\otimes n}$ means the $n$-fold tensor product, e.g. $V^{\otimes 2} = V\otimes V$. A tensor field is just a smoothly indexed family of tensors.

For simplicity, I'll just talk about the manifold $\mathbb{R}^n$, but anywhere I explicitly write out $\mathbb{R}^n$ (as opposed to $V$), you could just as well use a submanifold of $\mathbb{R}^n$, e.g. a 1-dimensional curve in $\mathbb{R}^n$. A rank-$k$ tensor field on $\mathbb{R}^n$ is a (suitably smooth) function $\tau : \mathbb{R}^n \to (V^{\otimes k} \multimap \mathbb R)$ where $V$ is itself $\mathbb{R}^n$. Now say we write the directional derivative of $\tau$ in some direction $v \in V$ at $x \in \mathbb{R}^n$ as $D(\tau)(x; v)$. The result itself would be a rank-$k$ tensor, i.e. $D(\tau)(x; v) : V^{\otimes k} \multimap \mathbb R$. So what is the type of $D(\tau)$ itself. Well we know the type of $\tau$ and we know the type of $V$ and we know $D(\tau)(x; v)$ is linear in $v$ and non-linear in $x$. So we have $$D(\tau) : \mathbb{R}^n \to (V \multimap (V^{\otimes k}\multimap\mathbb{R}))$$ but it is easy to show that $(V \multimap (V^{\otimes k}\multimap\mathbb{R})) \cong (V\otimes V^{\otimes k} \multimap \mathbb R) = (V^{\otimes k+1}\multimap \mathbb R)$. Which is to say $D(\tau)$ as a function of $x$ alone — essentially currying $D(\tau)$ — is a rank-$(k+1)$ tensor field.

So to answer your question, you find the gradient of a tensor field by viewing the directional derivative as a linear function of the direction. When you have a basis, as you do for $\mathbb{R}^n$, this linear function can be represented by a vector of partial derivatives (which are directional derivatives along the basis directions). A bit more concretely, $$D(\tau)(x)_i = \frac{\partial\tau}{\partial x_i}(x)$$

And yes, even high school, single variable calculus was doing this for the $n = 1$ case. It's just that $\mathbb{R}^{\otimes k} \cong \mathbb{R}$ for any $k$, and $(\mathbb{R}\multimap\mathbb{R}) \cong \mathbb{R}$.

$\endgroup$
  • 1
    $\begingroup$ I've never see this fancy symbol $\multimap$ before. Is there some mathematical theory in which this symbol is popular? Can you also write $D(\tau)$ for 2-tensor in matrix notation, so we can have some concrete case? $\endgroup$ – Fallen Apart Apr 22 '16 at 22:46
  • $\begingroup$ @FallenApart $\multimap$ is a commonly used symbol in linear logic for linear implication. The "linear" in "linear logic" means something rather different than the "linear" in "linear algebra", though it does turn out that there are connections that make this more than just a reuse of the symbol. $\endgroup$ – Derek Elkins Apr 22 '16 at 22:53
  • 1
    $\begingroup$ I also noticed that you treat rank-n tensors as covector fields. I.e. $\tau$ is a section of $\bigotimes T^*M.$ So you indentified vectors with covectors. If we are in $\mathbb{R}^n$ it can be done. However in arbitrary manifold we cannot define such derivative $D$ of arbitrary vector field (or tensor field). Im I right? $\endgroup$ – Fallen Apart Apr 22 '16 at 22:58
  • $\begingroup$ @FallenApart Given a suitable connection, the above can be generalized to an arbitrary "vector" bundle. The above would have worked just as well with $V^{\otimes k}\multimap\mathbb{R}$ replaced with some other "vector" space, though obviously there'd be no simplification. The quotes are because I prefer to work in synthetic differential geometry, which tends to be much nicer, where the analog of $\mathbb R$ is not a field, so you get modules instead of vector spaces. $\endgroup$ – Derek Elkins Apr 23 '16 at 1:35
  • $\begingroup$ @Derek Elkins Thanks for the answer. Is the notation you've used in your answer, the commonly accepted notation, as I haven't seen that notation used previously? $\endgroup$ – Perturbative Apr 23 '16 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.