I have just seen the tensor algebra for a Lie algebra. Let $\mathfrak{g}$ be a Lie algebra, and let $T(\mathfrak{g})$ be its tensor algebra.

I am told this is a Hopf algebra with coproduct $\Delta(x)=x\otimes 1 + 1\otimes x$. This confuses me.

The tensor algebra is $T(\mathfrak{g})=\Bbb C\oplus \mathfrak{g}\oplus (\mathfrak{g}\otimes \mathfrak{g})\oplus \cdots$.

I take it then, that $T(\mathfrak{g})\stackrel{\Delta}{\longrightarrow} T(\mathfrak{g})\otimes T(\mathfrak{g})$.

So that $x\otimes 1$ lives in $T^1(\mathfrak{g})\otimes T^0(\mathfrak{g})$ and $1\otimes x \in T^0(\mathfrak{g})\otimes T^1(\mathfrak{g})$.

But I must be careful:

  1. There is an isomorphism $(A\oplus B)\otimes C \cong (A\otimes C) \oplus (B\otimes C)$ and the same for left tensors. But this was for $R$-modules where $R$ is a commutative ring. Each of $\mathfrak{g}^{\otimes n}$ can be treated as $\Bbb C$-modules, so I suppose I can use this here?
  2. Above was finite though. Do I get a similar distributivity for the infinitely many terms in the direct sum case?
  3. Can I deduce that $T(\mathfrak{g})\otimes T(\mathfrak{g})= \displaystyle\bigoplus_{i+j=d, \,\, d\in \Bbb Z^+_0}( T^i(\mathfrak{g})\otimes T^j(\mathfrak{g}))$ (I mean, is this true, rather than; does it immediately follow from above)

We say $T(\mathfrak{g})$ is a 'Hopf-algebra' because it has mapps out of itself in particular ways. This doesn't mean we say $\Delta(x)$ is in the hopf algebra or anything right? It's simply that it has some nice structure regarding maps out of itself?

  • $\begingroup$ We say that it is a Hopf algebra because we have some extra maps (note that the comultiplication is not enough, we also need a counit and an antipode). Those maps are considered part of the "data" of being a Hopf algebra, in the same way that the multiplication is part of the "data" of being an algebra. $\endgroup$ – Tobias Kildetoft Sep 28 '17 at 9:17
  • $\begingroup$ @TobiasKildetoft Thanks for your comment. $\endgroup$ – user469689 Sep 28 '17 at 11:03

I think that you might be heading in the wrong direction to understand this definition, so I'll explain a way to interpret it. I won't use any facts about Lie algebras, so let $V$ be a vector space. I want to define a map $$ \Delta: T(V) \to T(V) \otimes T(V)$$ such that for all $x \in V$, $$ \Delta(x) = x \otimes 1 + 1 \otimes x$$ and somehow "extend this up to all of $T(V)$". Here is what is intended by this:

  1. If $A$, $B$ are algebras, then $A \otimes B$ becomes an algebra with the product $(a_1 \otimes b_1)(a_2 \otimes b_2) = a_1 a_2 \otimes b_1 b_2$.
  2. $T(V)$ is universal in the sense that any linear map $V \to A$ (where $V$ is a vector space and $A$ is an algebra) extends uniquely to a map of algebras $T(V) \to A$.
  3. Take the map $d: V \to T(V) \otimes T(V)$, $d(x) = x \otimes 1 + 1 \otimes x$ and extend it using point (2), where $T(V) \otimes T(V)$ is treated as an algebra via point (1).

After this, you need to check that $\Delta$ actually does what it should in the Hopf algebra: it plays well with the counit, is coassociative, etc.

These definitions are quite easy to use. For example, if $x, y \in V$, then we can find that

$$\begin{aligned} \Delta(xy) &= \Delta(x) \Delta(y) & \text{Since $\Delta$ is a map of algebras}\\ &= (x \otimes 1 + 1 \otimes x)(y \otimes 1 + 1 \otimes y) \\ &= xy \otimes 1 + x \otimes y + y \otimes x + 1 \otimes xy & \text{By algebra structure on $A \otimes B$} \end{aligned}$$

  • $\begingroup$ This clarifies many things that I have learned, even outside of this question. Thank you very much! $\endgroup$ – user469689 Sep 28 '17 at 10:22
  • $\begingroup$ So this sort of motivates why we suppress the tensor for elements of $T(\mathfrak{sl}_2)$ say. Since $$[(e\otimes f\otimes h)\otimes(e\otimes e)]\cdot [(e\otimes e) \otimes (f\otimes f)] = [efh\otimes e^2]\cdot [e^2\otimes f^2] = efhe^2\otimes e^2f^2$$ is much clearer. Since there are more or less, two different tensors to consider. $\endgroup$ – user469689 Sep 28 '17 at 10:37

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