Let $\mathfrak{g}$ denote a Lie algebra, $K$ a field, and view $K$ as a trivial $\mathfrak{g}$-module (that is, define $x \cdot a = 0$ for all $x \in \mathfrak{g}, a \in K$). In other words, we have a trivial structure map $\rho:\mathfrak{g} \rightarrow K$ sending everything to zero. By the universal property of the universal enveloping algebra $U\mathfrak{g}$, we can associate to this structure map a unique algebra homomorphism $\epsilon:U \mathfrak{g} \rightarrow K$, called the augmentation of $U\mathfrak{g}$.

I guess I must be confused regarding the construction of the induced map, because to me it seems that this map should just send everything to zero. For instance, on the generators of $U \mathfrak{g}$, shouldn't we have $\epsilon(x_1 \otimes \cdots \otimes x_n) = \rho(x_1) \cdots \rho(x_n) = 0$?

Thanks for your time.


The universal enveloping algebra $U\mathfrak{g}$ is a unital associative algebra. The map $\epsilon$ you've described above would send the identity element of this algebra to 1, and the elements of $\mathfrak{g}$ (which, along with the identity, generate $U\mathfrak{g}$ as an algebra) to zero. So it's not the zero map.

  • $\begingroup$ Ah, very nice! I totally forgot about the identity element. So, to recap, if we view $K$ as sitting inside $U \mathfrak{g}$, then $\epsilon$ will act as the identity on $K$, and send everything else to zero. $\endgroup$ – Alex Provost Apr 25 '13 at 17:18
  • $\begingroup$ Yes, if you're careful about what you mean by "everything else". One needs to pick a vector space complement to the $K$ sitting inside $U\mathfrak{g}$. This complement is precisely the subalgebra of $U\mathfrak{g}$ generated by $\mathfrak{g}$. This complement is sent to zero. $\endgroup$ – Alistair Savage Apr 25 '13 at 17:33

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