# Prove a metric is bilinear

1. Jun 30, 2015

### Simone Furcas

How could I proof this $ds^2=cos^2(v)du^2+dv^2$ is bilinear?

2. Jun 30, 2015

### Fredrik

Staff Emeritus
I don't even know what that is unless there's a metric involved. How do you define the the function that you want to prove is bilinear?

If this is a problem from a book or some kind of homework, we will need the full problem statement.

If it's just something you encountered in a book, then please tell us the name of the book, and where in the book you found it.

3. Jun 30, 2015

### Simone Furcas

I translate it, it's in Italian. I've just solved it by myself.
This is first fundamental form$ds^2=cos^2(v)du^2+dv^2$ v∈(-$\pi$/2,$\pi$/2). Check it is bilinear, symmetric and positive.

4. Jun 30, 2015

### Fredrik

Staff Emeritus
OK, if you solved it, then maybe you don't want to talk about it, but I'm curious how you made sense of the problem. A metric is bilinear by definition, and so is the first fundamental form. (I wasn't familiar with that term, but I just looked it up). So I don't see how to extract a function from your notation that needs to be proved to be bilinear.

If we're dealing with a metric, what you wrote can can be interpreted as a sloppy way of writing $g=g_{ij}dx^i \otimes dx^j$. This is a formula that holds because of the bilinearity of the metric: For all u,v, we have
$$g(u,v)=g(u^i e_i, v^je_j)=u^i v^j g(e_i,e_j) =u^i v^j g_{ij} =g_{ij} dx^i(u) dx^j(v) =g_{ij} dx^i\otimes dx^j (u,v).$$ This implies that $g=g_{ij}dx^i\otimes dx^j.$ How am I supposed to interpret what you wrote if not as $g_{uu}=\cos v$ and so on? I'm not sure that this interpretation makes sense if we don't already know that g is a metric, or at least that it's a bilinear form.

5. Jun 30, 2015

### Simone Furcas

I'll try answering by smartphone. It is an exercise to show the student remember that first fundamental form is a tensorial product addition. I red in my note my professor 's idea, he showed as you do the tensorial product, furthermore he takes 2 general vector of the space and with an unusual notation he wrote (du^2+dv^2)(X,Y) with X,Y variables, after that du^2(X,Y)+dv^2(X,Y) using tensorial,product it become du(X)du(Y)+dv(X)dv(Y) and after he wrote Xand y as sum of components xi and ei, ei is the I element of the base.. A friend of mine told me that the reason would be clear in a master's course of the same professor. I used this notation and I took. (x+Y,Z) as variables, and because it I obviously bilinear, after long and boring count finished. By phone I hard to answer, I hope being clear.