# Problem 2.1

The representation of this tensor, which maps pairs of covectors to numbers, follows the representation of covectors, where we use the set of vectors which are mapped to one, and the representation of metric tensors, where we use the set of vectors which are mapped to one. We use the set of covectors which are mapped to one by the tensor, and plot them as lines in the vector space. The representation will thus be the envelope of all these lines.

If we write the covector (finally we understand this notation) (careful, these L's may look like 1's on some screens)

k dx + l dy

then the tensor maps this covector to k^2 + 2 l^2. Thus we have

k^2 + 2 l^2 = 1

This looks just like a dispersion relation, albeit homogeneous of degree two (that is what makes it a tensor).

A single line representing one of these covectors will have the equation

k x + l ly = 1

To find the envelope, we want this line to intersect the line with slightly different k and l values, consistent with the above relation. This gives us another condition

- 2 l x + k y = 0

To derive that, think of k and l depending upon some parameter, and differentiate both the dispersion relation and the equation of the line with respect to this parameter and simplify.

Now if we square the above two equations and add twice the first to the second we find that we can eliminate both k and l, to find

2 x^2 + y^2 = 1.

Thus the envelope of the above dispersion relation, and the above tensor, is an ellipse.

# Problem 2.2

Same techniques as above. Now get equations

2 k l = 1

k x + l y = 1

-k x + l y = 0

And now square and subtract the two equations to find

2 x y = 1

as the equation of the envelope. This is a rectangular hyperbola.