# To see a tensor

1. Jun 6, 2008

### snoopies622

A vector is drawn as an arrow, a covector (one-form) as a series of parallel lines. Is there a way to pictorially represent a tensor of rank greater than one? I want to have an intuitive/geometric sense of what it means to parallel transport such an object, and without a picture I don’t have one.

2. Jun 7, 2008

### Phlogistonian

The outer product of two vectors is a simple tensor.

$$u^a v^b = w^{ab}$$

So for visualization purposes, you can imagine a tensor as a pair of arrows emanating from the same point.

3. Jun 7, 2008

### lbrits

The book Gravitation, by Misner, Thorne, Wheeler, discusses this ad nauseam. I recommend you take a look at that.

4. Jun 7, 2008

### tiny-tim

Mtw

Yeah … MTW really rocks on this!

5. Jun 7, 2008

### snoopies622

Thanks; I just happen to have that massive book on loan from the UNH physics library right now. I like Phlogistonian's idea, too. I guess if a (2,0) tensor can be imagined as a pair of arrows emanating from the same point then a (0,2) tensor like the metric tensor can be visualized as two overlapping sets of parallel lines that curve along with the coordinate system, although I suspect that there are limitations to such things...

6. Jun 13, 2008

### zpconn

I've never known much about visualizing tensors. Could someone explain how this visualization works? It's obvious that a vector can be thought of as an arrow, but I'm not sure how a covector would be a series of parallel lines to be honest.

One visualization I am familiar with applies only to differential n-forms. It grows out of integration theory. Basically, you think of an n-form as being the sort of "density" or volume element that you would integrate over a manifold.

7. Jun 13, 2008

### robphy

8. Jun 13, 2008

### lbrits

Well, not really parallel lines, but parallel surfaces. Think of the function $$f(x)$$. The gradient $$df$$ or $$\nabla f$$ defines a one-form, and if you contract with a vector $$\vec{v}$$, you get the directional derivative of $$f$$ in the direction pointed by $$\vec{v}$$.

If you take a curve with tangent vector $$\vec{v}(\lambda)$$ and you integrate $$\langle df, \vec{v} \rangle$$ along the curve, then by the fundamental theorem of calculus, you are integrating $$df/d\lambda$$, or how much f changes. Now think of surfaces $$f(x) = const$$, where each surface is evaluated for a different constant, and the constants are say, 1 unit apart. The integral you just computed tells you how many surfaces you have to cross as you move along the curve. MTW calls this "bongs of a bell" but anyway. So people visualize covectors, oneforms of the form $$df$$, as stacked surfaces, like layers of an onion.

9. Jun 13, 2008

### zpconn

Very nice. That's a neat way of doing it. But am I right in supposing it only works for exact forms (since you visualize it as the level surfaces of a function f such that the one-form is given by df)?

10. Jun 13, 2008

### lbrits

Yup.

11. Jun 13, 2008

### robphy

It can always work locally.

12. Jun 13, 2008

### lbrits

If said 1-form is not closed...? What is $$\omega = y\, dx$$ the differential of?

13. Jun 13, 2008

### Hurkyl

Staff Emeritus
I think he means pointwise (e.g. draw the pictures in the tangent space at the point of interest), rather than being perfectly accurate within an entire open neighborhood.

14. Jun 22, 2008

### MeJennifer

15. Jun 22, 2008

### robphy

Thanks.
Sorry... I don't have that with letter-size pages.
...but here is an early version: