Talk:Tensor algebra

Problem with Hopf Algebra
Under Coalgebra: Coproduct we have $$\Delta(k)=k(1\boxtimes 1)=k\boxtimes 1=1\boxtimes k$$ But under Hopf algebra: Compatibility we have $$(\nabla \circ (S \boxtimes \mathrm{id}) \circ \Delta)(k) = (\nabla \circ (S \boxtimes \mathrm{id})) (k\boxtimes k)$$ This can't be right. I'd think the second line is incorrect which implies the definiton of antipode for tensor algebras is wrong, namely we should have $$S(k) = -k$$ for $$k \in K$$. The compatibility condition then works out. I'm new to wikipedia so I don't know if I should just go ahead and edit? 0Amgine0 20:50, 4 April 2020 (UTC)

Untitled
I removed the following statement:
 * To have the complete algebra of tensors, contravariant as well as covariant, one should take T(W) where W is the direct sum of V and its dual space - this will consist of all tensors TIJ with upper indices J and lower indices I, in classical notation.

Classically, a tensor over V is an element of V⊗...⊗V⊗V*⊗...⊗V*; with the construction given above, one would also get mixed terms such as V⊗V*⊗V. To get the true algebra of all classical tensors, one would have to impose relations so that the elements of V commute with those of V*. 21:13, 5 Sep 2004 (UTC)

Ambiguous sentence about Quotients
This sentence may produce a misunderstanding: Because of the generality of the tensor algebra, many other algebras of interest are constructed by starting with the tensor algebra and then imposing certain relations on the generators, i.e. by constructing certain quotients of T(V). Examples of this are the exterior algebra, the symmetric algebra, other Schur functors, Clifford- and Weyl algebras and universal enveloping algebras.

From an historical standpoint, as far as I know, tensor algebra was defined by Ricci about 10 years after Clifford developed his algebra. In turn, 30 years before that, Grassmann had published his Extension Theory, which is the basis of both the modern exterior algebra (aka Grassmann algebra) and Clifford algebra. See http://modelingnts.la.asu.edu/html/evolution.html. The above inserted quoted text is ambiguous because the reader may interpret the words "are constructed by starting with" as related with the history of mathematics. Paolo.dL 16:13, 8 June 2007 (UTC)


 * Point taken, and I also feel that perhaps "are constructed" is a bit limiting (there are other constructions out there). How about "can be constructed"?  Also, I will link quotients to make it clear that we mean quotient algebras rather than some other sort of quotient.  Silly rabbit 16:21, 8 June 2007 (UTC)

Thank you, Silly Rabbit. My suggestion (changes in bold): Because of the generality of the tensor algebra, many other algebras of interest can be constructed by ... Examples of this are the Grassmann (= exterior)-  and symmetric algebras, Clifford- and Weyl algebras, universal enveloping algebras and Schur functors. However, historically tensor algebra was developed after Clifford algebra, which in turn was based on Grassmann's Extension theory.

The last sentence may need refinements. I am not an expert in this field, therefore I won't edit and I will leave the final decision to others. Paolo.dL 16:32, 8 June 2007 (UTC)

Fock space
Not even a remark on the link with Fock space? Link or should I say that it is identical? or maybe specialized to Hilbert spaces

In physics, Fock spaces describe states with undetermined number of particles, it is really surprising that the same construction appears. What is the reason?? — Preceding unsigned comment added by Noix07 (talk • contribs) 18:27, 15 February 2014 (UTC)


 * A Fock space is a kind of tensor algebra with extra conditions: the vector space is a Hilbert space and it is a direct sum of symmetrized or antisymmetrized tensor products of the Hilbert spaces. So they aren't identical. Would be worth a mention as an example. --Mark viking (talk) 19:16, 15 February 2014 (UTC)

Assessment comment
Substituted at 02:38, 5 May 2016 (UTC)

Bialgebra/Hopf algebra definitions can't be right.
The sections on the bialgebras cannot possible be right, as currently written. I can see glimmers of correctness, but... Here's one problem, this definition:
 * $$\Delta(v_1 \otimes \dots \otimes v_m ) := \sum_{i=0}^{m}

(v_1 \otimes \dots \otimes v_i) \otimes (v_{i+1} \otimes \dots \otimes v_m)$$ what does it mean for the index to run from i=0 ? Is the definition trying to define v0 to be 1 and ... !? Huh? Changing the index to i=1 just makes the right side equal to m times the left side, which is pointless. What was the idea here?

For m=1, I expect to get Δ : V → V ⊗ V = T2V, yeah? I don't seem how to plug in m=1 into the above, unless I use that weird v0=1 trick which is maybe OK, but seems to fail for m=2...

Next, this definition also has problems:
 * $$\Delta(x_1\otimes\dots\otimes x_m) = \sum_{p=0}^m \sum_{\sigma\in\mathrm{Sh}_{p,m-p}} \left(x_{\sigma(1)}\otimes\dots\otimes x_{\sigma(p)}\right)\otimes\left(x_{\sigma(p+1)}\otimes\dots\otimes x_{\sigma(m)}\right)$$

Again there is the bit with the sum p=0? Is it trying to say that $$x_{\sigma(0)}=1$$??? so as to recover the m=1 case as
 * $$\Delta(x)=x\otimes1+1\otimes x$$

That almost makes sense, but the m=2 case can't be right, the dimensions are wrong.

This also doesn't make sense:
 * $$\Delta(x_1\otimes\dots\otimes x_m) = \Delta(x_1)\Delta(x_2)\cdots\Delta(x_m).$$

lets examine the m=2 case:
 * $$\Delta(x_1\otimes x_2) = \Delta(x_1)\Delta(x_2).$$

what does the right hand even mean? Does it mean this?
 * $$\Delta(x_1\otimes x_2) = \Delta(x_1)\otimes\Delta(x_2)?$$

that cannot possibly be correct, it completely fails to satisfy the connecting axiom for a bialgebra, you have to swap the terms around, as you multiply. This is one of the basic axioms of a bialgebra, -- the very first one on the bialgebra page, you can't just blow it off.

There's some formula that looks kind-of-like this, but this is not it. 67.198.37.16 (talk) 04:20, 18 September 2016 (UTC)


 * I'm blanking that entire section. As written, its non-sense. 67.198.37.16 (talk) 05:24, 18 September 2016 (UTC)


 * OK I see what the problem is. The formulas were almost right, just need fixing and explanation. I'll fix it. 67.198.37.16 (talk) 13:34, 18 September 2016 (UTC)

Confusing universal property diagram
The diagram for the universal property doesn't match the diagram on the Universal property page; it should have the forgetful functor U applied to T(V) and A. I understand that this is probably meant to be implicit in the i and f maps of the diagram, but its inconsistent and a bit confusing. — Preceding unsigned comment added by Vapniks (talk • contribs) 10:11, 3 October 2018 (UTC)

TODO List
Here's a list of questions and answers this article could discuss:


 * Insert Fock space as an example, as mentioned above.
 * Remark also works for modules over commutative rings.
 * Remark: works whenever an iterated tensor product is possible. So e.g. for R-R-bimodules even when R is not commutative.
 * Define cocommutative and show that it holds.
 * Symmetric algebras are commutative and co-commutative. Group algebras are only commutative and co-commutative when the group is abelian. This is why symmetric algebras get used in cohomology.
 * What is the interplay between quotienting and the Hopf algebra construction? Clearly, the exterior and symmetric algebras endure. What about the Clifford and Weyl algebras?
 * Can one work with a braided vector space, when constructing the Hopf algebra, or not? Discuss consistency.
 * Can one start from here, and bridge the gap over to the Hopf algebras as used in chomology, or is bridging that gap too off-topic for this article? In particular, the cohomology use replaces each $$T^nV$$ by $$H^n$$ and although the Alexander-Whitney theorem tells you how to build tensor products of singular chains and the Kunneth formula and the universal coefficient theorem and the cup product and etc. but I'm unclear as to where obstructions may lie.
 * Tensor algebra functor as adjoint to forgetful functor to unitary associative algebras. symmetric algebra functor as adjoint to forgetful to commutative assoc algebras, etc.
 * Some less trivial explict examples: The free Lie algebra functor as the retract of the tensor algebra. More simply just the mapping of the commutator, or equally that the universal enveloping algebra inherits a Hopf algebra structure from the ensor algebra. Explain why "retract" is reasonable word to use.
 * The Milnor-Moore theorem in general.
 * The principle fiber bundle where the tensor algebra is the vertical space. Special cases. Zing back to e.g. Fock space.

All this needs mention or expansion. 67.198.37.16 (talk) 18:23, 19 September 2016 (UTC)

A suggestion for a small improvement
I think this article would be improved if the inclusion i : V &rarr; A were defined before it is used. Specifically it needs to be stated explicitly, not just implied, that i is used to identify V with T1V.96.11.154.131 (talk) 00:00, 1 August 2020 (UTC)

Definition of the coproduct
We are defining a coproduct, not a product, so why is there a link to shuffle product? Also, why is the sum taken over (p,m-p+1) shuffles? This doesn't make any sense to me, the number of terms isn't even right.--345Kai (talk) 11:32, 6 October 2021 (UTC)
 * There was clearly a typo (now fixed): (p,m-p+1) instead of (p+1,m-p). The shuffle product is involved for having a concise notation for the many terms involved in the expression of the coproduct. I agree that the explanations are rather confusing and could be improved. However, if one understands the definition of shuffle product, the notation for the formulas using it is almost self explanatory. D.Lazard (talk) 13:49, 6 October 2021 (UTC)