The whole debacle illustrates several major problems with the non-mainstream Right. They are:
1) Lack of a moral compass which allows malign elements to infiltrate the group.
2) High T, Low IQ membership which favours unthinking intuitive action.
3) A lack of an understanding of what it means to be Right.
4) A lack of an understanding of what we are up against.
Still, the events represent a strategic victory for the Dissident Right. And by Dissident Right, it is as Gottfried originally envisaged it. A Right that was built upon the traditions and identity of the West minus the modernistic ideologies trying to infiltrate it. The Charlottesville debacle seems to have pushed enough people to disavow themselves from the Natsocs which makes me think that future infiltration by them will be neutralised. They are now persona non grata. The Social Pathologist dissects it.
Moving on from first part.
Suppose ∇ is a derivative operator, and gab is a metric, on the manifold M. Then ∇ is compatible with gab iff ∇a gbc = 0.
Suppose γ is an arbitrary smooth curve with tangent field ξa and λa is an arbitrary smooth field on γ satisfying ξn∇nλa = 0. Then
ξn∇n(gabλaλb) = gabλaξn ∇nλb + gabλbξn ∇nλa + λaλbξn ∇ngab
Suppose first that ∇ngab = 0. Then it follows immediately that ξn∇ngabλaλb = 0. So ∇ is compatible with gab. Suppose next that ∇ is compatible with gab. Then ∀ choices of γ and λa (satisfying ξn∇nλa =0), we have λaλbξn∇ngab = 0. Since the choice of λa (at any particular point) is arbitrary and gab is symmetric, it follows that ξn∇ngab = 0. But this must be true for arbitrary ξa (at any particular point), and so we have ∇ngab = 0.
Note that the condition of compatibility is also equivalent to ∇agbc = 0. Hence,
0 = gbn∇aδcn = gbn∇a(gnrgrc) = gbngnr∇agrc + gbngrc∇agnr
= δbr∇agrc + gbngrc∇agnr = ∇agbc + gbngrc∇agnr.
So if ∇agbc = 0,it follows immediately that ∇agbc = 0. Conversely, if ∇agbc =0, then gbngrc∇agnr = 0. And therefore,
0 = gpbgscgbngrc∇agnr = δnpδrs∇agnr = ∇agps
The basic fact about compatible derivative operators is the following.
Suppose gab is a metric on the manifold M. Then there is a unique derivative operator on M that is compatible with gab.
It turns out that if a manifold admits a metric, then it necessarily satisfies the countable cover condition. And then it guarantees the existence of a derivative operator.) We do prove that if M admits a derivative operator ∇, then it admits exactly one ∇′ that is compatible with gab.
Every derivative operator ∇′ on M can be realized as ∇′ = (∇, Cabc), where Cabc is a smooth, symmetric field on M. Now
∇′agbc = ∇agbc + gnc Cnab + gbn Cnac = ∇agbc + Ccab + Cbac. So ∇′ will be compatible with gab (i.e., ∇′agbc = 0) iff
∇agbc = −Ccab − Cbac —– (1)
Thus it suffices for us to prove that there exists a unique smooth, symmetric field Cabc on M satisfying equation (1). To do so, we write equation (1) twice more after permuting the indices:
∇cgab = −Cbca − Cacb,
∇bgac = −Ccba − Cabc
If we subtract these two from the first equation, and use the fact that Cabc is symmetric in (b, c), we get
Cabc = 1/2 (∇agbc − ∇bgac − ∇cgab) —– (2)
Cabc = 1/2 gan (∇ngbc − ∇bgnc − ∇cgnb) —– (3)
This establishes uniqueness. But clearly the field Cabc defined by equation (3) is smooth, symmetric, and satisfies equation (1). So we have existence as well.
In the case of positive definite metrics, there is another way to capture the significance of compatibility of derivative operators with metrics. Suppose the metric gab on M is positive definite and γ : [s1, s2] → M is a smooth curve on M. We associate with γ a length
|γ| = ∫s1s2 gabξaξb ds,
where ξa is the tangent field to γ. This assigned length is invariant under reparametrization. For suppose σ : [t1, t2] → [s1, s2] is a diffeomorphism we shall write s = σ(t) and ξ′a is the tangent field of γ′ = γ ◦ σ : [t1, t2] → M. Then
ξ′a = ξads/dt
We may as well require that the reparametrization preserve the orientation of the original curve – i.e., require that σ (t1) = s1 and σ (t2) = s2. In this case, ds/dt > 0 everywhere. (Only small changes are needed if we allow the reparametrization to reverse the orientation of the curve. In that case, ds/dt < 0 everywhere.) It
|γ’| = ∫t1t2 (gabξ′aξ′b)1/2 dt = ∫t1t2 (gabξaξb)1/2 ds/dt
= ∫s1s2 (gabξaξb)1/2 ds = |γ|
Let us say that γ : I → M is a curve from p to q if I is of the form [s1, s2], p = γ(s1), and q = γ(s2). In this (positive definite) case, we take the distance from p to q to be
d(p,q)=g.l.b. |γ|:γ is a smooth curve from p to q.
Further, we say that a curve γ : I → M is minimal if, for all s ∈ I, ∃ an ε > 0 such that, for all s1, s2 ∈ I with s1 ≤ s ≤ s2, if s2 − s1 < ε and if γ′ = γ|[s1, s2] (the restriction of γ to [s1, s2]), then |γ′| = d(γ(s1), γ(s2)) . Intuitively, minimal curves are “locally shortest curves.” Certainly they need not be “shortest curves” outright. (Consider, for example, two points on the “equator” of a two-sphere that are not antipodal to one another. An equatorial curve running from one to the other the “long way” qualifies as a minimal curve.)
One can characterize the unique derivative operator compatible with a positive definite metric gab in terms of the latter’s associated minimal curves. But in doing so, one has to pay attention to parametrization.
Let us say that a smooth curve γ : I → M with tangent field ξa is parametrized by arc length if ∀ ξa, gabξaξb = 1. In this case, if I = [s1, s2], then
|γ| = ∫s1s2 (gabξaξb)1/2 ds = ∫s1s2 1.ds = s2 – s1
Any non-trivial smooth curve can always be reparametrized by arc length.