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Tensor algebras are the most general algebras you can get (categorically inclined people would call them a “free object” in a suitable category), but because of

by joppy 6y ago

Tensor algebras are the most general algebras you can get (categorically inclined people would call them a “free object” in a suitable category), but because of this they are almost never useful on their own. We pretty much always study specialisations of tensor algebras, of which GA (or more generally Clifford algebras) are an example. There are no interesting formulas in a tensor algebra, because the whole idea of it being “most general” (every other algebra is a quotient) is that it has no relations. It is only when passing to more specific algebras (symmetric, exterior, GA, ...) that the subject really becomes interesting.

creata 6y ago
Yep. For a silly analogy, it's like saying that set theory will replace every field of mathematics.
pjbk 6y ago
> There are no interesting formulas in a tensor algebra, because the whole idea of it being “most general” (every other algebra is a quotient) is that it has no relations. It is only when passing to more specific algebras (symmetric, exterior, GA, ...) that the subject really becomes interesting. And yet you can still go a level further, with objects like those in Moon & Spencer's holor theory that include generalizations for transformation rules beyond covariant or contravariant, generalized connections similar --for example-- to universal Christoffel symbols, and effectively creating rules for tensors, pseudotensors and stranger array-like objects.
- JadeNB 6y ago
  > And yet you can still go a level further, with objects like those in Moon & Spencer's holor theory that include generalizations for transformation rules beyond covariant or contravariant, generalized connections similar --for example-- to universal Christoffel symbols, and effectively creating rules for tensors, pseudotensors and stranger array-like objects. One can always go farther, but I'm not sure that this is doing that. I looked up the Wiki page https://en.wikipedia.org/wiki/Parry_Moon#Holors https://en.wikipedia.org/wiki/Parry_Moon#Holors and found: > But this is incorrect, in general, for the definition of a tensor includes a specific dependence on coordinate transformation. This is very much a physicist's definition of a tensor; a mathematician's definition includes no such thing (although one may prove results about co-ordinate transformations as a theorem). For example, there is no such thing, mathematically, as an intrinsically covariant or contravariant tensor, although, having chosen a fixed vector space $V$, one may certainly put a bi-grading on $T(V \otimes V^*)$ that makes that sort of distinction.

by joppy 6y ago

creata 6y ago

Yep. For a silly analogy, it's like saying that set theory will replace every field of mathematics.

pjbk 6y ago

> There are no interesting formulas in a tensor algebra, because the whole idea of it being “most general” (every other algebra is a quotient) is that it has no relations. It is only when passing to more specific algebras (symmetric, exterior, GA, ...) that the subject really becomes interesting. And yet you can still go a level further, with objects like those in Moon & Spencer's holor theory that include generalizations for transformation rules beyond covariant or contravariant, generalized connections similar --for example-- to universal Christoffel symbols, and effectively creating rules for tensors, pseudotensors and stranger array-like objects.

JadeNB 6y ago
> And yet you can still go a level further, with objects like those in Moon & Spencer's holor theory that include generalizations for transformation rules beyond covariant or contravariant, generalized connections similar --for example-- to universal Christoffel symbols, and effectively creating rules for tensors, pseudotensors and stranger array-like objects. One can always go farther, but I'm not sure that this is doing that. I looked up the Wiki page https://en.wikipedia.org/wiki/Parry_Moon#Holors https://en.wikipedia.org/wiki/Parry_Moon#Holors and found: > But this is incorrect, in general, for the definition of a tensor includes a specific dependence on coordinate transformation. This is very much a physicist's definition of a tensor; a mathematician's definition includes no such thing (although one may prove results about co-ordinate transformations as a theorem). For example, there is no such thing, mathematically, as an intrinsically covariant or contravariant tensor, although, having chosen a fixed vector space $V$, one may certainly put a bi-grading on $T(V \otimes V^*)$ that makes that sort of distinction.