Ciro Santilli @cirosantilli 34

The definition of the "dot product" of a general space varies quite a lot with different contexts.

Most definitions tend to be bilinear forms.

We use the unqualified generally refers to the dot product of Real coordinate spaces, which is a positive definite symmetric bilinear form. Other important examples include:The rest of this section is about the ${\mathbb{R}}^n$ case.

The positive definite part of the definition likely comes in because we are so familiar with metric spaces, which requires a positive norm in the norm induced by an inner product.

The default Euclidean space definition, we use the matrix representation of a symmetric bilinear form as the identity matrix, e.g. in ${\mathbb{R}}^3$:so that:

A form is a function from a vector space to elements of the underlying field of the vector space.

Examples:

Generalization of orthogonal group to preserve different bilinear forms. Important because the Lorentz group is $SO(1,3)$.

The group of all transformations that preserve some bilinear form, notable examples:

From effect of a change of basis on the matrix of a bilinear form, remember that a change of basis $C$ modifies the matrix representation of a bilinear form as:

So, by taking $S=C^T$, we understand that two matrices being congruent means that they can both correspond to the same bilinear form in different bases.

As usual, it is useful to think about how a bilinear form looks like in terms of vectors and matrices.

Unlike a linear form, which was a vector, because it has two inputs, the bilinear form is represented by a matrix $M$ which encodes the value for each possible pair of basis vectors.

In terms of that matrix, the form $B(x,y)$ is then given by:

Multivariate polynomial where each term has degree 2, e.g.:is a quadratic form because each term has degree 2:but e.g.:is not because the term $x^3$ has degree 3.

More generally for any number of variables it can be written as:

There is a 1-to-1 relationship between quadratic forms and symmetric bilinear forms. In matrix representation, this can be written as:where $\vec{x}$ contains each of the variabes of the form, e.g. for 2 variables:

Strictly speaking, the associated bilinear form would not need to be a symmetric bilinear form, at least for the real numbers or complex numbers which are commutative. E.g.:But that same matrix could also be written in symmetric form as:so why not I guess, its simpler/more restricted.

Skew-symmetric bilinear map that is also a bilinear form.

The main interest of this theorem is in classifying the indefinite orthogonal groups, which in turn is fundamental because the Lorentz group is an indefinite orthogonal groups, see: all indefinite orthogonal groups of matrices of equal metric signature are isomorphic.

It also tells us that a change of basis does not the alter the metric signature of a bilinear form, see matrix congruence can be seen as the change of basis of a bilinear form.

The theorem states that the number of 0, 1 and -1 in the metric signature is the same for two symmetric matrices that are congruent matrices.

For example, consider:

The eigenvalues of $A$ are $1$ and $4$, and the associated eigenvectors are:symPy code:and from the eigendecomposition of a real symmetric matrix we know that:

Now, instead of $P$, we could use $PE$, where $E$ is an arbitrary diagonal matrix of type:With this, would reach a new matrix $B$:Therefore, with this congruence, we are able to multiply the eigenvalues of $A$ by any positive number $e_1^2$ and $e_2^2$. Since we are multiplying by two arbitrary positive numbers, we cannot change the signs of the original eigenvalues, and so the metric signature is maintained, but respecting that any value can be reached.

Note that the matrix congruence relation looks a bit like the eigendecomposition of a matrix:but note that $D$ does not have to contain eigenvalues, unlike the eigendecomposition of a matrix. This is because here $S$ is not fixed to having eigenvectors in its columns.

But because the matrix is symmetric however, we could always choose $S$ to actually diagonalize as mentioned at eigendecomposition of a real symmetric matrix. Therefore, the metric signature can be seen directly from eigenvalues.

Also, because $D$ is a diagonal matrix, and thus symmetric, it must be that:

What this does represent, is a general change of basis that maintains the matrix a symmetric matrix.

Related:

symmetric bilinear maps that is also a bilinear form.

Intuition, please? Example? mathoverflow.net/questions/278641/intuition-for-symplectic-groups The key motivation seems to be related to Hamiltonian mechanics. The two arguments of the bilinear form correspond to each set of variables in Hamiltonian mechanics: the generalized positions and generalized momentums, which appear in the same number each.

Seems to be set of matrices that preserve a skew-symmetric bilinear form, which is comparable to the orthogonal group, which preserves a symmetric bilinear form. More precisely, the orthogonal group has:and its generalization the indefinite orthogonal group has:where S is symmetric. So for the symplectic group we have matrices Y such as:where A is antisymmetric. This is explained at: www.ucl.ac.uk/~ucahad0/7302_handout_13.pdf They also explain there that unlike as in the analogous orthogonal group, that definition ends up excluding determinant -1 automatically.

Therefore, just like the special orthogonal group, the symplectic group is also a subgroup of the special linear group.

We looking at the definition the orthogonal group is the group of all matrices that preserve the dot product, we notice that the dot product is one example of positive definite symmetric bilinear form, which in turn can also be represented by a matrix as shown at: Section "Matrix representation of a symmetric bilinear form".

By looking at this more general point of view, we could ask ourselves what happens to the group if instead of the dot product we took a more general bilinear form, e.g.:The answers to those questions are given by the Sylvester's law of inertia at Section "All indefinite orthogonal groups of matrices of equal metric signature are isomorphic".