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
   

Confusion in Black Holes

I've been trying to make my way through Majid's paper, ``Almost commutative Riemannian geometry: wave operators'', particularly the section where he constructs the model for a Schwarzschild black hole. I've not had much success. I'm meant to reconstruct the model over a $ {\mathbb{F}_p}$, but I'm stuck on basic definitions. I'll discuss two of those things in this blog, first the vector bundle (aka projective module) and then the Grassmann connection.

1. NCG Vector Bundle for The Schwarzschild Solution

To construct the model, we start by reconsidering our notion of 3-dimensional space. Rather than thinking of coordinates $ {(x_1, x_2, x_3)}$, we're going to recast ``space'' as a ``coordinate algebra'', in particular, an algebra of polynomials $ {k[x_1, x_2, x_3]}$ over a field $ {k}$ (we'll let $ {k = \mathbb{R}}$ for now, but my task is to redo this section of the paper with $ {k=\mathbb{F}_p}$. Moreover, we're working in a sphere, so we also request that our algebra contain functions rational in $ {r}$, where $ {r^2 = x_1^2 + x_2^2 + x_3^2}$. Hence our ``space'' is the algebra $ {A = k[x_1, x_2, x_3, r, r^{-1}]}$ modded out by the aforementioned relation.

For such an NCG space (nevermind that $ {A}$ is actually commutative here), we define a vector bundle as a projective module. An easy way to get a projective module is to take a few copies of $ {A}$ under the image of an idempotent $ {E \in M_n(A)}$, e.g, let $ {E}$ be such an idempotent, then $ {\mathcal{E} = Im(E)}$ is our vector bundle. In this case we're taking the 3 by 3 matrix:

$ \displaystyle E = \begin{pmatrix} 1 - \frac{x_1^2}{r^2} & - \frac{x_1 x_2}{r^2} & - \frac{x_1 x_3}{r^2} \\ - \frac{x_2 x_1}{r^2} & 1 - \frac{x_2^2}{r^2} & - \frac{x_2 x_3}{r^2} \\ - \frac{x_3 x_1}{r^2} & - \frac{x_3 x_2}{r^2} & 1 - \frac{x_3^2}{r^2} \end{pmatrix}$

Thus our vector bundle is the subspace $ {\mathcal{E} = Im(E) \subset A^3}$. We expect elements of this vector bundle to be ``3-vectors'' with entries from $ {A}$. Yet in the paper, Majid states that $ {\omega_i = \text{d}x_i - \frac{x_i \text{d}r}{r}}$ for $ {i = 1, \, 2, \, 3}$ spans the 2-dimensional bundle $ {\mathcal{E}}$. But (judging by the $ {\text{d}x_i}$ and $ {\text{d}r}$ terms) each $ {\omega_i}$ is in our bimodule of 1-forms $ {\Omega_1}$. Where am I going wrong?

2. Grassmann Connections in NCG

Let's assume I'm not hopelessly confused about that vector bundle thing. Recall that a connection on an NCG vector bundle is a linear map $ {\nabla_{\mathcal{E}}: \mathcal{E} \rightarrow \Omega_1 \otimes \mathcal{E}}$ that obeys the following rule:

$ \displaystyle \nabla_{\mathcal{E}} (a s) = \text{d} a \otimes s + a\nabla_{\mathcal{E}}(s) \; \forall a \in A \; s \in \mathcal{E} $

According to a proposition I've read in Majid's lecture notes, if $ {\mathcal{E} = Im(E)}$ for a projector $ {E \in M_n(A)}$, then we have a ``Grassmann connection''

$ \displaystyle \nabla_{\mathcal{E}} (E v) = E \text{d}(Ev) = E\left(\text{d}(E)v + E(\text{d}v)\right) = E\text{d}(E)v + E(\text{d}v)$

Where $ {v \in A^n}$ and $ {\text{d}}$ acts on $ {v}$ and $ {E}$ component-wise. But the image of the connection is suppose to live in $ {\Omega_1 \otimes \mathcal{E}}$. $ {\text{d}v}$ lives in $ {\Omega_1^n}$, and $ {\text{d}E}$ lives in $ {M_n(\Omega_1)}$. How do we get to our tensor product space $ {\Omega_1 \otimes \mathcal{E}}$ ?

3. Sources

1. Section 4 of Almost commutative Riemannian geometry: wave operators