representation on the tensor algebra over a Lie algebra I'm now reading <Differential Geometry, Lie Groups, and Symmetric Spaces> by Helgason.
In Proposition 1.1 of Chapter 2, he says:
If $\rho$ is a representation $\mathfrak{a} \to \mathfrak{gl}(V)$, then there is a unique representation $\tilde{\rho} : T(\mathfrak{a}) \to \mathfrak{gl}(V)$ 
such that $\tilde{\rho}(X) = \rho (X)$ for all $X \in \mathfrak{a}$, where $\mathfrak{a}$ is a Lie algebra and $T(\mathfrak{a})$ is the tensor algebra over $\mathfrak{a}$.
But there is no explanation about the existence or the uniqueness of the representation $\tilde{\rho}$.
Since the tensor algebra $T(\mathfrak{a})$ is the direct sum
$T(\mathfrak{a}) = \bigoplus_{k=0}^{\infty} T^k \mathfrak{a}$, I think we can consider $\tilde{\rho}$ as an extension of $\rho$. 
To define such extension on $T(\mathfrak{a})$, we need to define an algebra homomorphism on each $k$-th tensor of $\mathfrak{a}$. 
And then I got lost in the middle, since it should be a representation on $V$. 
I know I can do tensor products of representations, but then it would result a representation on  $T^k V$.
How should I understand this?
 A: The elements of, say, $T^2\mathfrak{a}$ are linear combinations of elementary tensors $a\otimes a'$, $a,a'\in\mathfrak{a}$. Such a tensor will act on a vector $v\in V$ by the rule
$$(a\otimes a')\cdot v=a\cdot (a'\cdot v)$$
as it must for otherwise the action would not match with the multiplication of the tensor algebra. The same rule is applied to longer elementary tensors:
$$(a_1\otimes a_2\otimes\cdots\otimes a_m)\cdot v=a_1\cdot(a_2\cdot(\cdots (a_m\cdot v)\cdots).$$
This leads to a representation of $T(\mathfrak{a})$ by the universal property of the tensor algebra. Any linear transformation $\phi$ from a vector space $W$ to an associative algebra $A$ (both over the same field) gives rise to a unique homomorphism of algebras $T(W)\to A$ extending $\phi$. Here $W$ takes the role of $\mathfrak{a}$ and $\mathfrak{gl}(V)$ plays the part of $A$.
Observe that we don't need $W$ to be a Lie algebra for this to work.

Lie algebra people use the so called universal enveloping algebra instead of the tensor algebra. It is a quotient algebra of $T(\mathfrak{a})$ gotten by imposing the relations $a\otimes a'-a'\otimes a=[a,a']$, so identifying certain elements (="commutators") of $T^2(\mathfrak{a})$ with elements of $\mathfrak{a}$. We also impose all consequences of those relations so that the associative law holds. The way to do that is to mod out the ideal $I$ generated by those relations, and we end up with $U(\mathfrak{a})=T(\mathfrak{a})/I$.
Then, assuming that $\rho:\mathfrak{a}\to\mathfrak{gl}(V)$ is a representation of a Lie algebra, the ideal $I$ is automatically mapped to zero (because all its generators do), and the homomorphism $T(\mathfrak{a})\to \mathfrak{gl}(V)$ we got from the universal property induces a homomorphism of associative algebras $U(\mathfrak{a})\to \mathfrak{gl}(V)$ extending the original representation.
