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Michi’s blog » Reading Merkulov: Differential geometry for an algebraist (2 in a series)

 Reading Merkulov: Differential geometry for an algebraist (2 in a series)

  • February 11th, 2006
  • 11:03 pm

So, in the last installment, we got to know smooth manifolds and charts, atlases and some nice topological tricks and tweaks. For this round, we follow Merkulov onward, and pretty soon stumble across category theory and sheaves. The notes I’m following here are from the link on Merkulov’s website. It starts, however, with a nice discussion of temperatures in archipelagos. Go read it – I imagine I’m almost comprehensible at that part of the text.

A map from a subset of a smooth manifold to \mathbb R is called a smooth function on the subset if for every x in the subset and a coordinate chart at x, the n-to-1 variable function f\circ\phi^{-1} is smooth at the point \phi(x).

What this means is that if we have some point on our manifold, we can take out our chart over that part of the manifold and check out the function of our coordinates as chained through the chart and the function from the manifold portion. This can be made even more clear by taking a look at coastal temperatures. Suppose you’re on a boat in the Stockholm archipelago. You have a location on the earth (which is, for our purposes, a manifold). There is some function that to each point on earth assigns a temperature. This is our function. Now, for the area around Vaxholm, you can take out your chart, and you get coordinates in terms of something like “135 mm north and 25 mm west on this sheet of my atlas!”. So we have a set of coordinates: (135,25). These coordinates correspond to a point on the manifold, namely your current position. You can break out your thermometer and measure the temperature here, and thus you get the function value at this position. As you move about around Vaxholm, you get different temperatures, so with this fixed choice of a chart over your area, you really do get a function \mathbb R^2\to\mathbb R. The function is smooth if regardless of where you are or what chart you’re using, this function you get is a smooth function.

The set of all smooth functions on the smooth manifold M form an algebra over \mathbb R, which we denote by \mathcal E^\infty(M). Indeed, if f is a smooth function, then of course \lambda f is a smooth function for real constants f. And f+g is smooth if f and g both are smooth. The same holds for the pointwise product of two smooth functions. So the axioms check out.

Analogously the complex analytic or holomorphic functions on a subset of a complex manifold are defined; whereby a complex manifold is a real manifold of even dimension, with \mathbb R^{2n} identified with \mathbb C^n and the analytic structure comes from requiring all transformation functions to be complex analytic. It’s either a 2n-dimensional manifold or an n-dimensional complex manifold, depending on how you look at it.

Now, note that smooth and holomorphic behave very different from each other. All holomorphic functions on a compact complex manifold are constant; and even if you have a holomorphic function vanishing on an open subset of its domain, it vanishes everywhere; but for \mathbb R^n there are non-trivial smooth functions that vanish outside an arbitrary open ball.

A map between two manifolds is said to be smooth if for any pair of relevant coordinate charts, the map from \mathbb R^n to \mathbb R^n by going from the coordinates up to the first manifold, with the map to the second and then back to coordinates ends up being smooth regardless of choice of charts.

A smooth map of manifolds \psi\colon M_1\to M_2 induces in an ordinary manner the pullback map of rings of smooth functions
\psi_*\colon\mathcal E^\infty(M_2)\to\mathcal E^\infty(M_1) by \psi_*(f)=f\circ\psi.

Note that every continuous map between smooth manifolds is homotopy equivalent to a smooth map. Thus, when computing homotopy of a manifold it is enough to work with equivalence classes of smooth maps S^m\to M, which tends to simplify things.

Graded algebraic structures

Most interesting algebraic structures can be equipped with a grading. This decomposes the structure into a direct sum of part-structures where each partstructure is a representative of the structure in question; and where each partstructure contains only elements of the same degree – called homogenous. For instance, any polynomial ring is a graded structure, by taking the degree of the monomial as grading.

The elements of \Hom(A, B) for graded structures are graded as well in a natural way – a map being homogenous of degree i if |f(v)|=|v|+i.

For structures with multiplication – i.e. anything more advanced than modules/vector spaces – we add a sign to the multiplication according to the Koszul sign convention: any term that interchanges the homogenous vectors v and w gets the factor (-1)^{|v||w|}.

Some examples

A \mathbb Z-graded Lie algebra is a \mathbb Z-graded vector space equipped with a special morphism [ , ]\in\Hom(V\otimes V,V) such that
[v_1,v_2]=-(-1)^{|v_1||v_2|}[v_2,v_1] and
[[v_1,v_2],v_3]+[[v_2,v_3],v_1]+[[v_3,v_1],v_2]=0
for all homogenous v_1, v_2, v_3 in V.

A \mathbb Z-graded commutative algebra is a \mathbb Z-graded vector space equipped with a special morphism \mu\in\Hom(V\otimes V,V) such that
\mu(v_1,v_2)=-(-1)^{|v_1||v_2|}\mu(v_2,v_1) and
\mu(v_1,\mu(v_2,v_3))=\mu(\mu(v_1,v_2),v_3)
for all homogenous v_1, v_2, v_3 in V. We tend to write v_1v_2 for \mu(v_1,v_2).

The symmetric tensor algebra \odot V on a \mathbb Z-graded vector space is the quotient of the tensor algebra \otimes V=\bigoplus_{n=0}^\infty \otimes^nV by the ideal generated by all expressions on the form v_1\otimes v_2-(-1)^{|v_1||v_2|}v_2\otimes v_1. This is a graded commutative algebra.

The skew-symmetric tensor algebra \wedge V on a \mathbb Z-graded vector space is the quotient of the tensor algebra \otimes V=\bigoplus_{n=0}^\infty \otimes^nV by the ideal generated by all expressions on the form v_1\otimes v_2+(-1)^{|v_1||v_2|}v_2\otimes v_1. This is a graded commutative algebra.

Note that \odot V\cong\wedge V[1]. More precisely \odot^n V=(\wedge^nV[1])[-n] as graded vector spaces.

Exercise

Given a \mathbb Z-graded vector space V show that the structure of a \mathbb Z-graded Lie algebra on V[1] is equivalent to the following data: A degree -1 element [\bullet]\in\Hom(\otimes^2V,V) satisfying
[v_1\bullet v_2]=(-1)^{|v_1||v_2|+|v_1|+|v_2|}[v_2\bullet v_1] and the Jacobi identity.

Proof

We take the hint in the notes and let s:V\to V[1] be the natural isomorphism of degree 1, and let [v_1\bullet v_2]=s^{-1}[sv_1,sv_2]. If v has degree n, then v lives in degree n-1 in V[1], and so sv has degree n-1. So [sv_1,sv_2] has degree |v_1|+|v_2|-2 and |s^-1[sv_1,sv_2]|=|v_1|+|v_2|-1.
As our next step, we examine [v_1\bullet v_2] and its relationship to [v_2\bullet v_1]. Thus
[v_1\bullet v_2]=s^{-1}[sv_1,sv_2]=s^{-1}(-(-1)^{|sv_1||sv_2|}[sv_2,sv_1])
and now note that |sv_1||sv_2|=(|v_1|+1)(|v_2|+1)=|v_1||v_2|+|v_1|+|v_2|+1 so the sign (which other than that just tunnels out through s^{-1}) will end up yielding the expression
(-1)^{|v_1||v_2|+|v_1|+|v_2|}s^{-1}[sv_2,sv_1]
which is just
(-1)^{|v_1||v_2|+|v_1|+|v_2|}[v_2\bullet v_1].

The Jacobian identity is the same. So if we have a Lie structure on V[1] we will get the described structure on V. We can go through basically the same moves as we did to show that if we have this structure, we can define a Lie bracket on V[1] which, in endeffect, will yield precisely a Lie structure based in these axioms.

It turns out that precisely this structure occurs if we take V=k[x^1,\dots,x^n,\psi_1,\dots,\psi_n] and let |\psi_a|=|x^a|+1. The bracket would be defined as
[f\bullet g]=(-1)^{|f|}\delta(fg)-(-1)^{|f|}(\Delta f)g-f\Delta g
where
\Delta=\sum_{a=1}^n \frac{\partial^2}{\partial x^a\partial \psi_a}
This bracket can be verified to satisfy the given axioms, and thus actually defines the structure of graded Lie algebra on V[1].

For the next installment, we start with section 3, which discusses categories. Of course, categories are well known to the person writing this, and thus I will not talk much about them, but skip on to the meaty stuff behind them.

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