TITLE: What's the use of quadratic forms?
QUESTION [14 upvotes]: Starting with the abstract concept of a vector space, I can see why we'd want to add some structure to be able to perform useful operations.  For instance if we add a metric/ norm to a vector space we can talk about distances.  If we add an inner product to a vector space we can talk about angles.  These two operations also give us a bunch of inequalities (like Cauchy-Schwarz) that we get as well.
But I don't see the point of equipping our vector space with some degree two polynomial.  What does that get us?  Is there some geometric meaning to it (like how we got distance and angle from norm and inner product)?

REPLY [2 votes]: Another place where bilinear and quadratic forms appear naturally and have geometric meaning is in algebraic and differential topology. If $M$ is a compact connected oriented $2n$ dimensional manifold, the wedge product induces a bilinear form on the vector space $H_{\mathrm{dR}}^n(M)$ of $n$-dimensional de Rham cohomology classes on $M$. Using Poincaré duality, one can interpret this bilinear form as computing generic signed intersections between $n$ dimensional submanifolds of $M$.
I can demonstrate this intuitively for the case $M$ is a two-dimensional torus. Consider the following image (taken from  the Wolfram Mathworld page on homology intersection):

For the torus, $H_{\mathrm{dr}}^1(M)$ is a two dimensional real vector space and we can choose a basis $\mathcal{B} = (v_1, v_2)$ for $H_{\mathrm{dr}}^1(M)$ under which $v_1$ corresponds to the blue circle in the picture and $v_2$ corresponds to the red circle. On $H_{\mathrm{dr}}^1(M)$ we have a bilinear form $g$ which encodes the intersections between the circles. We have $ g(v_1, v_1) = 0 $ which corresponds to the fact that if we perturb the blue circle a little, the intersection between the original blue circle and the perturbed circle is zero. Similarly, we have $g(v_2, v_2) = 0$. However, we (can choose the basis so that) have $g(v_1, v_2) = 1$ which corresponds to the fact that the blue circle and the red circle intersect at a single point (counted with a plus sign) and even if we perturb them a little, they will still generically intersect at a single point. If we change the order of $v_1$ and $v_2$, this doesn't effect the geometric intersection but does change the sign and so $g(v_2, v_1) = -1$. Thus, the intersection form $g$ is a bilinear form on $H_{\mathrm{dr}}^1(M)$ represented by the matrix
$$ [g]_{\mathcal{B}} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. $$
In this case, $g$ is an anti-symmetric bilinear form but if the dimension of $M$ is divisible by $4$, it will be a symmetric bilinear form (corresponding to a quadratic form). The form $g$ encodes information about submanifolds sitting inside $M$ and is an important invariant of $M$. It can be used to show that two manifolds are not homeomorphic but showing that corresponding bilinear forms are not equivalent. You can read a bit more about this here.