what have you done to me?

And the answer is, we don't.

The unhoused people who come here seeking refuge 99% of the time understand that they will be kicked out if they misbehave.

The people you have to watch out for are Jessica, who only came because the kid she didn't want had to visit for a homework assignment and she just *needs* to yell at her child for asking to borrow two books or stay an extra five minutes, or Michael, who came in to look at porn on our computers for whatever fucking reason, or Karen who just wanted to come by to throw a fit that the particular book she wanted was checked out and harrass our staff about our collection being too limited.

And we did what we could to help. He was incredibly kind and patient despite his obvious anxiety and stress, more than most able bodied, housed patrons are to us under much less dire conditions. I liked knowing that we were the first place he came.

We have so many people like this who come in everyday. Many are quiet and keep to themselves, but sometimes they talk to us.

They tell us about how they're taking a few courses on a scholarship they applied for from our library's computer at the local community college to get their diploma. Or ask about a manga or dvd or book we might have to help them pass the time.

One woman, who comes in daily with her tattered walker always says hello to me and likes to work on the new jigsaw puzzle with me when we set one out.

So like, treat unhoused people like people. Treat disabled people like people. I don't want my library to feel like the only safe space in the world, but I'm glad it can be one of them.

local matchmaker losing it over yet another unrealistic demand from the clientele 🤣🤣

This is what's called a poetic epithet, and it happens all the time in writing. Consider, for example, Homer and his wine-dark sea, or Shakespeare and his death-mark'd love.

If you were then to start using them instead of the word they're describing, you'd be making a kenning. For example, if instead of describing the sun as honey-bright, you said "the honey-drop in the sky". Kennings are best known from Norse writings, like "the path of serpents" for gold, or "Grímnir's lip-streams" for poetry itself.

◡ – / ◡ – / ◡ – / ◡ – / ◡ – The sun / was ho/ney-bright / that af/ternoon

"pants are for SQUARES"

In the last two parts, we defined (co)homology groups of (co)chain complexes and extended this to define (co)homology groups of topological spaces, producing algebraic invariants for topological structures.

This was all explained fairly abstractly, so as an example, in this post I will explain and define de Rham cohomology.

In short, de Rham cohomology assigns to every smooth manifold the cochain complex of vector spaces of differential forms of increasing degree with the exterior derivative operator as boundary operator. The de Rham cohomology groups of a smooth manifold are then the cohomology groups of this so-called de Rham cochain complex.

My aim is that by the end of this post, you will be able to understand this definition.

This post assumes familiarity with the definition of a smooth n-manifold and a covariant tensor field.

Differential k-Forms

A differential k-form should be intuitively thought of as some (degree k homogeneous) expression that can be integrated over an oriented k-manifold.

A differential 1-form for instance can be integrated over an interval. A 2-form can be integrated over a surface. And so on.

Examples include f(x)dx + g(y)dy and h(x, y)dx∧dy and the like.

Formally, a differential k-form ω on a smooth n-manifold M is a totally antisymmetric (0, k)-tensor field over M.

Differential k-forms being tensor fields should make a lot of sense!

For example, given a differential 1-form f’(x)dx, this can be written as the gradient function df(x), which we will properly define later but can be thought of intuitively for now. The gradient function df(x) takes a tangent vector as an input at each point and linearly maps it to a number describing the directional derivative of f(x) in that direction at that point. This is exactly the definition of a (0, 1)-tensor field.

Similarly, differential 2-forms can be thought of as taking two tangent vectors at each point and giving the bidirectional derivative of a binary function at that point.

Differential k-forms being totally antisymmetric (meaning if you swap any of their arguments, the resulting number is negated) also makes sense, since integration over a manifold with the opposite orientation will then give a negated result.

It is also worth mentioning at this point that the differential k-forms on a smooth manifold M for some fixed k form a real vector space denoted Ωᵏ(M). It is these vector spaces that will form the cochains of our de Rham complex.

Smooth scalar fields (aka functions) on a smooth manifold are interpreted as differential 0-forms.

Differential forms of degree n and m can also be multiplied via the exterior product ∧ to give a differential (n+m)-form. For 0-forms, this just corresponds to ordinary multiplication of functions.

The Exterior Derivative

Just as differential k-forms can be integrated over oriented k-manifolds to give real numbers, they can also be differentiated to give differential (k+1)-forms.

This is done through the exterior derivative, denoted d.

While d is technically defined separately for each M and k, the same symbol d is used for every M and k.

The exterior derivative is defined as the unique (collection of) linear map(s) d: Ωᵏ(M) -> Ωᵏ⁺¹(M) for all k satisfying the following:

For k = 0, d maps a differential 0-form (function) f to its gradient function df, which takes a tangent vector at each point on M and gives the directional derivative of f at each point in the direction of the tangent vector at that point

d(α∧β) = dα∧β + (-1)ᵏα∧dβ, where k is the degree of α

d² = 0

The first point expresses the base case while the second point expresses a generalised product rule called the graded product rule.

The third point is harder to give an intuitive explanation of, but I’m sure there are people better able to explain it than me. It ultimately boils down to the fact that “the boundary of a boundary is zero”.

Ironically, the third point is actually the most important to emphasise, since this means that the image of the first application of d is a subset of the kernel of the second application of d, which is exactly what is required for the collection Ωᵏ(M) to form a cochain complex under d!

The de Rham Cochain Complex and de Rham cohomology

The de Rham cohomology theory hence associates to every smooth manifold M the following cochain complex:

This is thus known as the de Rham complex of M.

The de Rham cohomology groups of M are hence the cohomology groups of the de Rham complex of M.

Final Discussion

The de Rham cohomology theory can be reformulated in terms of the Eilenberg-Steenrod axioms by defining C(M) to be the de Rham complex of M and defining Hⁿ(X, A) to be precisely Hⁿ(C(X)/C(A)).

The nth de Rham cohomology group of a topological space X is then Hⁿ(X) = Hⁿ(X, Ø).

It can be checked that this indeed satisfies Eilenberg-Steenrod. In fact, de Rham cohomology relates strongly to another very popular cohomology theory called singular cohomology.

In particular, the de Rham cohomology groups of a smooth manifold M are always isomorphic to the cohomology groups of M attained from singular cohomology with real coefficients.

This means that de Rham cohomology and singular cohomology with real coefficients are essentially the same!

This is a powerful theorem called the de Rham theorem.