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Dacyn7mo10

-"this is just the lie algebra, and is why elements of it are always invertible."

First of all, how did we move from talking about numbers to talking about Lie algebras? What is the Lie group here? The only way I can make sense of your statement is if you are considering the case of a Lie subgroup of GL(n,R) for some n, and letting 1 denote the identity matrix (rather than the number 1) [1]. But then...

Shouldn't the Lie algebra be the monad of 0, rather than the monad of 1? Because usually Lie algebras are defined in terms of being equipped with two operations, addition and the Lie bracket. But neither the sum nor the Lie bracket of two elements of the monad of 1 are in the monad of 1.

[1] Well, I suppose you could be considering just the special case n=1, in which case 1 the identity matrix and 1 the number are the same thing. But then why bother talking about Lie algebras? The group is commutative, so the formalism does not appear to be necessary.

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by Alok Singh
1 min read2nd Jan 20238 comments

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Logic & Mathematics World Modeling
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just like how infinitesimals are useful because they're indiscernible from 0, but have the advantage of being able to be divided by.

the other important number besides 0 is 1. gpt even suggested it as i was typing it.

the monad of 1 is numbers of the form (1 + ε) where ε is an infinitesimal. multiplying by them should be almost as good as multiplying by 1, aka a freebie.

this is just the lie algebra, and is why elements of it are always invertible. by using the infinitesimal version of a clifford algebra on the tangent space, we can use its rotor represention to see how the exponential map gives lie's third theorem.

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[-]Alexander Gietelink Oldenziel7mo50

With all due respect with your brand as LessWrong's ornamental hermeneutic I'm afraid I'll need some clarification.

What is the monad of 1 exactly? A monad is a functor - what category are we talking about here?

In particular - what are the unit and multiplication maps?

(my guess:  for the unit and  but now I'm using nilsquare infinitesimals instead of invertible infinitesimals.)

 

I'm not sure what tangent space we are talking about - but I assume it's a Lie group (hyperfinite graph?) and we are looking at the tangent space of the identity element. In this case - what is the clifford algebra of the tangent space? Construction of a clifford algebra needs a choice of inner product (or more generally a quadratic form) - what are we picking here?

[-]Alok Singh7mo10

On any finite dim space we have a canon inner product by taking the positive definite one.

Monad is a synonym for infinitesimal neighborhood, common on the literature. Not the category theory monad.

Also hermeneutic lmfao

[-]Dacyn7mo30

-"On any finite dim space we have a canon inner product by taking the positive definite one."

What? A finite dimensional space has more than one positive definite inner product (well, unless it is zero-dimensional), this choice is certainly not canonical. For example in R^2 any ellipse centered at the origin corresponds to a positive definite inner product.

[-]Alok Singh6mo10

I was thinking the one corresponding to a unit circle, just the ordinary dot product.

Canon is probably the wrong word in a mathy context.

[-]Alok Singh7mo10

Also yes the infinitesimal neighborhood of the identity.

[-]Dacyn7mo10

-"this is just the lie algebra, and is why elements of it are always invertible."

First of all, how did we move from talking about numbers to talking about Lie algebras? What is the Lie group here? The only way I can make sense of your statement is if you are considering the case of a Lie subgroup of GL(n,R) for some n, and letting 1 denote the identity matrix (rather than the number 1) [1]. But then...

Shouldn't the Lie algebra be the monad of 0, rather than the monad of 1? Because usually Lie algebras are defined in terms of being equipped with two operations, addition and the Lie bracket. But neither the sum nor the Lie bracket of two elements of the monad of 1 are in the monad of 1.

[1] Well, I suppose you could be considering just the special case n=1, in which case 1 the identity matrix and 1 the number are the same thing. But then why bother talking about Lie algebras? The group is commutative, so the formalism does not appear to be necessary.

[-]TAG7mo10

Consider capitalising "lie".

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