A Portal Special Presentation- Geometric Unity: A First Look: Difference between revisions

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===== Choosing All Metrics =====
===== Choosing All Metrics =====


<p>[01:10:36] So we allow $$U^14$$ to equal the space of metrics on $$X^4$$ pointwise. Therefore, if we propagate on top of this, let me call this the projection operator. If we propagate on $$U^14$$ we are in some sense following a Feynman like idea of propagating over the space of all metrics, but not at a field level.
<p>[01:10:36] So we allow $$U^14$$ to equal the space of metrics on $$X^4$$ pointwise. Therefore, if we propagate on top of this, let me call this the projection operator. If we propagate on $$U^14$$ we are, in some sense, following a Feynman-like idea of propagating over the space of all metrics, but not at a field level.


<p>[01:10:58] At a pointwise tensorial level.
At a pointwise tensorial level.


<p>[01:11:03] Is there a metric on $$U^14$$? Well, we both want one and don't want one. If we had a metric from the space of all metrics, we could define fermions, but we would also lock out any ability to do dynamics. We want some choice over what this metric is, but we don't want full choice because we want enough to be able to define the matter fields to begin with.
<p>[01:11:03] Is there a metric on $$U^14$$? Well, we both want one and don't want one. If we had a metric from the space of all metrics, we could define fermions, but we would also lock out any ability to do dynamics. We want some choice over what this metric is, but we don't want full choice, because we want enough to be able to define the matter fields to begin with.


<p>[01:11:22] It turns out. That if this is $$X^4$$ and this is this particular endogenous choice of $$U^14$$ we have a 10 dimensional metric along the fibers. So we have a $$G^{10}_{\mu nu}$$. Further for every point in the fibers, we get a metric downstairs on the base space. So if we pull back the cotangent bundle, we get a metric $$G^4_{\mu \nu}$$ on $$\Pi^*$$ of the cotangent bundle of X. We now define the chimeric bundle, right? And the chimeric bundle is this direct sum of the vertical tangent bundle along the fibers with the pullback, which we're going to call the horizontal bundle from the base space. So the chimeric bundle is going to be the vertical tangent space of 10 dimensions to U direct sum, the four dimensional cotangent space which we're going to call horizontal to U. And the great thing about the chimeric bundle is, is that it has an a priori metric. It's got a metric on the four a metric on the 10 and we can always decide that the two of them are naturally perpendicular to each other.
<p>[01:11:22] It turns out if this is $$X^4$$ and this is this particular endogenous choice of $$U^14$$, we have a 10-dimensional metric along the fibers. So we have a $$G^{10}_{\mu nu}$$.  


<p>[01:13:00] Furthermore, it is almost canonically isomorphic to the tangent bundle or the cotangent bundle, because we either have 4 out of 14 or 10 out of 14 dimensions on the nose. So the question is, what are we missing. And the answer is that we're missing exactly the data of a connection. So this bundle, chimeric, C. We have C is equal to the tangent bundle of U up to a choice of a connection theta.
Further for every point in the fibers, we get a metric downstairs on the base space. So if we pull back the cotangent bundle, we get a metric $$G^4_{\mu \nu}$$ on $$\Pi^*$$ of the cotangent bundle of X.
 
We now define the chimeric bundle, right? And the chimeric bundle is this direct sum of the vertical tangent bundle along the fibers with the pullback, which we're going to call the horizontal bundle from the base space.
 
So the chimeric bundle is going to be the vertical tangent space of 10-dimensions to U direct sum the four-dimensional cotangent space, which we're going to call horizontal to U. And the great thing about the chimeric bundle is that it has an a priori metric. It's got a metric on the four, a metric on the 10, and we can always decide that the two of them are naturally perpendicular to each other.
 
<p>[01:13:00] Furthermore, it is almost canonically isomorphic to the tangent bundle or the cotangent bundle, because we either have 4 out of 14, or 10 out of 14 dimensions, on the nose. So the question is, what are we missing? And the answer is that we're missing exactly the data of a connection. So this bundle, chimeric $$C$$. We have $$C$$ equal to the tangent bundle of $$U$$ up to a choice of a connection $$\theta$$.


<p>[01:13:32] And this is exactly what we wanted. We have a situation where we have some field on the manifold X in the form of a connection which is amenable, more friendly to quantization, which is now determining a metric turning around the Levi-Civita game. And the only problem is, is that we've had to buy ourselves into a different space than the one we thought we wanted to work on.
<p>[01:13:32] And this is exactly what we wanted. We have a situation where we have some field on the manifold X in the form of a connection which is amenable, more friendly to quantization, which is now determining a metric turning around the Levi-Civita game. And the only problem is, is that we've had to buy ourselves into a different space than the one we thought we wanted to work on.
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