What is a sesquilinear form?

Question

Can anyone please make me understand in easy language what is a sesquilinear form? How could it be used on vector spaces over the field of complex numbers?

My Knowledge : Basic knowledge of vector space.

Answer 1

A sesquilinear form on a complex vector space is more or less like an inner product, where you take two vectors, and you produce a complex number as an output. For some applications the standard inner product (the one you probably learned for real vector spaces in high school) has a major drawback: the product of one vector with itself isn't necessarily a real number. That means that it isn't as useful as one might hope to talk about the lengths of vectors, or for decomposing a vector into a linear combination of given vectors, where the inner product works nicely in the real case.

Enter the standard sesquilinear form. What it does (in the standard basis, at least) is that it takes your two vectors, complex conjugates the components of the first one, then uses the standard inner product on the resulting two vectors. Now the product of one vector with itself is necessarily a (non-negative) real number, and lengths work nicely, as does decomposition.

As an example, say we have the complex vector $(1, 1)\in \Bbb C^2$, and we want to express it as a linear combination of $(i, 0)$ and $(0, i)$. Clearly the answer is $$ (1, 1) = -i(i, 0) -i(0,i) $$ but how do we get hold of those $-i$'s in general? In the real case, you would take the inner product of $(1, 1)$ with each of the two given vectors (because they're clearly orthogonal and unit length) and use that as the components. However, here it would give you $i$, not $-i$. And this time I've used orthogonal unit vetors. It gets even worse in the general case.

Using the standard sesquilinear form instead, we get $$ \langle(i, 0), (1, 1)\rangle =\overline{(i, 0)}\cdot (1, 1)\\ = (-i, 0)\cdot (1, 1) = -i $$ as we wanted (where $\cdot$ here denotes the standard inner product). We get $\langle(0, i), (1, 1)\rangle = -i$ in a similar manner (note that it's important that the basis vector comes first and the vector which is to be decomposed comes second). The general case also mimics the real case closely using the sesquilinear form (again, as long as you do the product in the correct order).

The standard sesquilinear form is used extensively in physics, particularly in quantum mechanics (where they call it "the inner product"). Specifically in quantum mechanics, the sesquilinear form is so important that they write their vectors as ${}\mid w\rangle$, anticipating a "conjugated" basis element $\langle v\mid{}$ so that you can put them together to make the form $\langle v\mid w\rangle$ and calculate the "inner product". The ${}\mid w\rangle$ is called a "ket" and $\langle v\mid{}$ is called a "bra", and when you put them together, you get a "bra-ket", or a bracket, which is $\langle v\mid w\rangle$.

Answer 2

A sesquilinear form is basically just an inner product with a little twist. In linear algebra we have inner products $\langle x,y\rangle$ (like the dot product $x\cdot y$) that take in two vectors and output a real number. By definition inner products have some useful properties:

$$ \begin{align*} \langle x,x\rangle & \ge 0 \\ \langle x,y\rangle & = \langle y,x\rangle \\ \langle x, \alpha y + z\rangle & = \alpha\langle x,y\rangle + \langle x,z\rangle \end{align*} $$

where $\alpha$ is just a real number. This kind of inner product is called a bilinear form.

One reason we like inner products is because they allow us to define norms if we let $\|x\| = \sqrt{\langle x, x\rangle}$. Norms give us a concept of size and distance on our vector space which can be used for all kinds of things.

But what if we are allowed to multiple $x$ by a complex number, or if we allow $\langle x,y\rangle$ to output a complex number? Then having a bilinear form isn't good enough: the norm of a vector has to be a real number, and if $\langle x,x\rangle$ is complex then $\|x\| = \sqrt{\langle x,x\rangle}$ will be complex too. It turns out that if we want to have $\langle x,x\rangle$ be real for all $x$, we need to have

$$ \langle x,y\rangle = \overline{\langle y,x\rangle} $$

where $\overline{\langle y,x\rangle}$ is your complex conjugate, i.e. $\overline{a + ib} = a - ib$. When we add this property our inner product becomes known as a sesquilinear form, because we can show that

$$ \begin{align*} \langle x,\alpha y + z\rangle & = \alpha\langle x,y\rangle + \langle x,z\rangle \\ \langle \alpha x + z, y\rangle & = \overline{\alpha}\langle x,y\rangle + \langle z,y\rangle \end{align*} $$

This is different from what we'd get with a bilinear form, where we'd have the same property but replace $\overline{\alpha}$ with $\alpha$.