Ciro Santilli @cirosantilli 34

Analogous to a linear form, a bilinear form is a Bilinear map where the image is the underlying field of the vector space, e.g. ${\mathbb{R}}^n\times {\mathbb{R}}^m\to \mathbb{R}$.

Some definitions require both of the input spaces to be the same, e.g. ${\mathbb{R}}^n\times {\mathbb{R}}^n\to \mathbb{R}$, but it doesn't make much different in general.

The most important example of a bilinear form is the dot product. It is only defined if both the input spaces are the same.

Linear map of two variables.

More formally, given 3 vector spaces X, Y, Z over a single field, a bilinear map is a function from:that is linear on the first two arguments from X and Y, i.e.:Note that the definition only makes sense if all three vector spaces are over the same field, because linearity can mix up each of them.

The most important example by far is the dot product from ${\mathbb{R}}^n\times {\mathbb{R}}^n\to \mathbb{R}$, which is more specifically also a symmetric bilinear form.

Notation used in quantum mechanics.

Ket is just a vector. Though generally in the context of quantum mechanics, this is an infinite dimensional vector in a Hilbert space like $L^2$.

Bra is just the dual vector corresponding to a ket, or in other words projection linear operator, i.e. a linear function which can act on a given vector and returns a single complex number. Also known as... dot product.

For example:is basically a fancy way of saying:that is: we are taking the projection of $y$ along the $x$ direction. Note that in the ordinary dot product notation however, we don't differentiate as clearly what is a vector and what is an operator, while the bra-ket notation makes it clear.

The projection operator is completely specified by the vector that we are projecting it on. This is why the bracket notation makes sense.

It also has the merit of clearly differentiating vectors from operators. E.g. it is not very clear in $x\cdot y$ that $x$ is an operator and $y$ is a vector, except due to the relative position to the dot. This is especially bad when we start manipulating operators by themselves without vectors.

This notation is widely used in quantum mechanics because calculating the probability of getting a certain outcome for an experiment is calculated by taking the projection of a state on one an eigenvalue basis vector as explained at: Section "Mathematical formulation of quantum mechanics".

Making the projection operator "look like a thing" (the bra) is nice because we can add and multiply them much like we can for vectors (they also form a vector space), e.g.:just means taking the projection along the $x+y$ direction.

Ciro Santilli thinks that this notation is a bit over-engineered. Notably the bra's are just vectors, which we should just write as usual with $\overrightarrow{v}$... the bra thing makes it look scarier than it needs to be. And then we should just find a different notation for the projection part.

Maybe Dirac chose it because of the appeal of the women's piece of clothing: bra, in an irresistible call from British humour.

But in any case, alas, we are now stuck with it.

This section is about the definition of the dot product over ${\mathbb{C}}^n$, which extends the definition of the dot product over ${\mathbb{R}}^n$.

Some motivation is discussed at: math.stackexchange.com/questions/2459814/what-is-the-dot-product-of-complex-vectors/4300169#4300169

The complex dot product is defined as:

E.g. in ${\mathbb{C}}^1$:

We can see therefore that this is a form, and a positive definite because:

Just like the usual dot product, this will be a positive definite symmetric bilinear form by definition.

Given a matrix $A$ with metric signature containing $m$ positive and $n$ negative entries, the indefinite orthogonal group is the set of all matrices that preserve the associated bilinear form, i.e.:Note that if $A=I$, we just have the standard dot product, and that subcase corresponds to the following definition of the orthogonal group: Section "The orthogonal group is the group of all matrices that preserve the dot product".

As shown at all indefinite orthogonal groups of matrices of equal metric signature are isomorphic, due to the Sylvester's law of inertia, only the metric signature of $A$ matters. E.g., if we take two different matrices with the same metric signature such as:and:both produce isomorphic spaces. So it is customary to just always pick the matrix with only +1 and -1 as entries.

where:

Remember that $\psi$ is a 4-vetor, gamma matrices are 4x4 matrices, so the whole thing comes down to a dot product of two 4-vectors, with a modified $\psi$ by matrix multiplication/derivatives, and the result is a scalar, as expected for a Lagrangian.

Like any other Lagrangian, you can then recover the Dirac equation, which is the corresponding equations of motion, by applying the Euler-Lagrange equation to the Lagrangian.

Appears to be analogous to the dot product, but also defined for infinite dimensions.

${\mathbb{R}}^4$ with a weird dot product-like operation called the Minkowski inner product.

Because the Minkowski inner product product is not positive definite, the norm induced by an inner product is a norm, and the space is not a metric space strictly speaking.

The name given to this type of space is a pseudometric space.

www.youtube.com/watch?v=tq7sb3toTww&list=PLxBAVPVHJPcrNrcEBKbqC_ykiVqfxZgNl&index=19 mentions that it is a bit like a dot product but for a tangent vector to a manifold: it measures how much that vector derives along a given direction.

The dot product is a positive definite matrix, and so we see that those will have an important link to familiar geometry.

Subcase of symmetric multilinear map:

Requires the two inputs $x$ and $y$ to be in the same vector space of course.

The most important example is the dot product, which is also a positive definite symmetric bilinear form.

When viewed as matrices, it is the group of all matrices that preserve the dot product, i.e.:This implies that it also preserves important geometric notions such as norm (intuitively: distance between two points) and angles.

This is perhaps the best "default definition".

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".