## Monday, November 28, 2011

### Climbing Out of Black Holes

Last month I published a two questions that had me stuck inside Majid's almost commutative black hole''. We'll start today be answering those questions while trying to avoid insulting myself and my readers, and then move on to calculate the Grassmann connection on our Black Hole Algebra.

## 1. NCG Vector Bundles For the Schwarzschild Solution

Our space'' is the algebra ${A = k[x_1, x_2, x_3, r, r^{-1}]}$, our vector bundle'' projective module ${\mathcal{E}}$ is the image of the operator given by the idempotent matrix ${P \in \text{M}_3(A)}$ with ${(P)_{ij} = \delta_{ij} - \frac{x_i x_j}{r^2}}$, and our calculus of 1-forms ${\Omega_1}$ is the bimodule generated by symbols ${\text{d}x_i}$ for ${i = 1, 2, 3}$. The question was how can ${\mathcal{E} \subset A^{\oplus 3}}$ by spanned by ${\omega_i = \text{d}x_i - \frac{x_i \text{d}r}{r}}$ for ${i = 1, 2, 3}$.
The key here is to realize that the ${\Omega_1}$ is isomorphic to ${A^{\oplus 3}}$. Consider ${A^{\oplus 3}}$ to be the space of row 3-vectors ${v = (a_1 \; a_2 \; a_3)}$, and give it the standard basis'' ${e_i = (\delta_{1 i} \; \delta_{2 i} \; \delta_{3 i} )}$. In this case, the idempotent acts on the right, and its action on the standard basis is ${e_i \cdot P = P_{ik}e_k}$. Under the isomorphism given by ${e_i \mapsto \text{d}x_i}$, we see that ${\mathcal{E}}$ is spaned by ${e_i \cdot P \mapsto P_{ik}\text{d}x_k = \text{d}x_i - \frac{x_i \text{d}r}{r} = \omega_i}$ as desired.

## 2. Grassmann Connections in NCG

Last time we defined a connection on an NCG vector bundle ${\mathcal{E}}$ as a linear map ${\nabla:\mathcal{E} \rightarrow \Omega_1 \otimes_{A} \mathcal{E}}$ that obeys certain conditions. We also defined the Grassmann Connection'' a map ${\nabla_P = p \circ \text{d}}$, where ${P}$ is the idempotent of the projective module. More explicitly, the Grassmann connection is, for a ${v \in A^{\oplus n}}$

$\displaystyle \nabla_P (v \cdot P) = \text{d} ( v \cdot P ) \cdot P = v \cdot \text{d} (P) \cdot P + \text{d}v \cdot P$

Where ${\text{d}}$ acts on ${v}$ and ${P}$ component-wise. We had asked how this definition makes sense, as its image appears to lie in ${\Omega_1^{\oplus n}}$.
My chief problem here was that I don't understand tensor products over a ring (or over an algebra, in this case). The key out of this problem is to realize two facts about tensor product spaces: Given module (over ${A}$) ${M}$, ${N}$, and ${R}$,

1. ${(M \oplus N) \otimes R}$ is isomorphic to ${M \otimes R \oplus N \otimes R}$, by ${(m, n) \otimes r \mapsto (m \otimes r, n \otimes r)}$, and
2. ${M \otimes A}$ is isomorphic to ${M}$ by ${m \otimes a \mapsto ma}$.

Given these two facts, it's not hard to see that ${\Omega_1^{\oplus n} \cong (\Omega_1 \otimes A)^{\oplus n} \cong \Omega_1 \otimes A^{\oplus n}}$. Then it's easy to see that the map ${P \circ \text{d}}$ maps to the right space.

## 3. The Grassmann Connection for our Black Hole Model

We'll calculate the action of the Grassmann Connection on our spanning set ${\omega_i}$. In this section, we'll identify (by the previously defined isomorphims) the standard basis'' ${e_i}$ with ${\text{d}x_i}$, so that ${\omega_i = e_i \cdot P = P_{ik}\text{d}x_k = \text{d}x_i - \frac{x \text{d}r}{r}}$. From the above definition, we have:

$\displaystyle \nabla_P (\omega_i) = \text{d}e_i \cdot P + e_i \text{d}P \cdot P$

${\text{d}e_i = 0}$, as ${e_i}$ is constant (or equivalently, because ${\text{d}^2=0}$). So the bulk of the problem lies in calculating the matrix product ${\text{d}P \cdot P}$. Let's write it out:

$\displaystyle (\text{d}P \cdot P)_{ij} = (\text{d}P)_{ik} P_{kj} = -\text{d}\left( \frac{x_i x_k}{r^2} \right) \left( \delta_{kj} - \frac{x_k x_j}{r^2} \right) = -\frac{x_i}{r^2} P_{kj}\text{d}x_k$

So that ${e_i \text{d}P \cdot P = -\frac{x_i}{r^2} P_{kl} P_{lj} dx_k \otimes dx_j = -\frac{x_i}{r^2} \omega_l \otimes \omega_l}$. Here I used the fact that P is idempotent and the generators of our algebra commute. This gives us our connection:

$\displaystyle \nabla_P (\omega_i) = -\frac{x_i}{r^2} \omega_l \otimes \omega_l$

## 4. Next Steps

We won't actually be using this connection much. Rather, we're going to define a new one based on the idea that ${\nabla (\text{d}x_i)}$ should be ${0}$. From here, we'll begin discussing NCG metrics, and write down one for our black hole.

## 5. Sources

1. Section 4 of Almost commutative Riemannian geometry: wave operators
2. Email conversations with Majid
3. Atiyah and MacDonald, Introduction to Commutative Algebra