I am confused in the Universal enveloping algebra associated to a Lie algebra $L$. Basically my doubt is in the product structure for universal enveloping algebra. By definition it is $T(L)/I$ where $T(L)$ is the tensor algebra and $I$ is the ideal of $T(L)$ generated by elements of the form $x \otimes y-y \otimes x-[xy]$. The product structure in $T(L)$ is given by extending the below product linearly.

$(x_1 \otimes x_2\otimes \ldots \otimes x_r)$.$(y_1 \otimes y_2\otimes \ldots \otimes y_s)=x_1 \otimes x_2\otimes \ldots \otimes x_r \otimes y_1 \otimes \ldots \otimes y_{s}$. Am I right? This will give a product structure in the quotient also. Then if $\{x_1, \ldots,x_n\}$ is a basis for $L$ then if $y_1, \ldots, y_n$ are the images of these $x_i$'s in the quotient $T(L)/I$ , then the monomials $y_1^{a_1}y_2^{a_2}\ldots y_n^{a_n}$ will form a basis for $T(L)/I$ by PBW basis theorem. My doubt is what is the product of these $y_i$'s say $y_1y_2$? What I understood is it is the image in the quotient of $x_1 \otimes x_2$. Am I correct?

If I am right how can we describe the universal enveloping algebra of $sl(2,\mathbb{C})$.? With respect to the standard basis $\{x,y,h\}$ by PBW basis theorem the monomials $x^a y^b h^c $ will form a basis. What is this power of the matrix here say $x^a$? Please help me. I am totally confused of these.

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    $\begingroup$ The product has nothing to do with the matrix product, so I am not really sure what sort of answer you are looking for. You are correct about the first part. $\endgroup$ – Tobias Kildetoft Jan 15 '17 at 14:44
  • $\begingroup$ So how can we describe the universal enveloping algebra of $sl(2,\mathbb{C})$? I want to prove that $1-x$ is not invertible in $U(sl(2,\mathbb{C})$. How can I proceed? $\endgroup$ – Anupam Ah Jan 15 '17 at 15:45
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    $\begingroup$ The first part is way broader than the last. I have no idea what sort of description would satisfy you for the first, but for showing that an element is not invertible, you should probably just play around with the relations (there is also a way using the Hopf algebra structure, but that has probably not been introduced to you). $\endgroup$ – Tobias Kildetoft Jan 15 '17 at 15:55
  • $\begingroup$ Actually, forget what I said about the Hopf algebra structure. I made a mistake there. $\endgroup$ – Tobias Kildetoft Jan 15 '17 at 16:41
  • $\begingroup$ Thank you.. I have one more doubt. What is the product say $x^2$? Is there any way to compute explicitly since we know what is x. $\endgroup$ – Anupam Ah Jan 16 '17 at 4:32

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