2024-11-07

This following is a very brief summary of what happened in class on 2024-11-07.

We began by recalling the construction of the tensor algebra, $\mathcal{T}(M)$. We noted that it actually has the structure of a graded ring, where the degree $k$ component is the $k^{\text{th}}$ tensor power, ${\mathcal{T}}^k(M)$. In particular, the degree 0 and 1 components provide $R$-module inclusions $R\hookrightarrow \mathcal{T}(M)$ and $M\hookrightarrow \mathcal{T}(M)$. We also described bases for $\mathcal{T}(M)$ in the case $R=F$ is a field and $M=V$ is an $n$-dimensional vector space.

We then roughly sketched out the bijection

First suppose you have an $R$-algebra morphism $\varphi :\mathcal{T}(M)\to A$. Forgetting that multiplicative structure, we then have $R$-module morphisms from each summand, ${\varphi }_k:{\mathcal{T}}^k(M)\to U(A)$. In particular, for $k=1$ we have an $R$-module morphism ${\varphi }_1:M\to U(A)$.

Conversely, suppose you have an $R$-module morphism ${\varphi }_1:M\to U(A)$. There is only one possible way to define $R$-module morphisms ${\varphi }_k:{\mathcal{T}}^k(M)\to U(A)$ such that the corresponding morphism $\varphi :\mathcal{T}\to U(A)$ is actually an $R$-algebra morphism, i.e., compatible with multiplication. For $k=0$, the $R$-module morphism ${\varphi }_0:R\to U(A)$ is entirely determined by the image of $1_R$. For $\varphi :\mathcal{T}(M)\to U(A)$ to be a ring morphism, we must have $1_R\mapsto 1_A$. Thus, ${\varphi }_0$ is completely determined. For $k=1$, our component function must agree with ${\varphi }_1$. For $k=2$, if $\varphi$ is to be a ring morphism then for every simple tensor $m\otimes m^{\prime }\in {\mathcal{T}}^2(M)$ we must have ${\varphi }_2(m\otimes m^{\prime })=\varphi (m\otimes m^{\prime })=\varphi (m)\varphi (m^{\prime })={\varphi }_1(m){\varphi }_1(m^{\prime })$. So, ${\varphi }_2$ is completely determined by ${\varphi }_1$, and so on. Finally, one can verify that the map $\varphi :\mathcal{T}(M)\to A$ defined in this way actually does define an $R$-algebra morphism, and these two associations are indeed mutual inverses.

We then proceeded to ask for a construction analogous to the tensor algebra construction, but which produced commutative $R$-algebras. This led us to the symmetric algebra functor. The construction was simple, namely quotienting the tensor algebra $\mathcal{T}(M)$ by the (graded) ideal $\mathcal{C}(M)$ generated by all tensors of the form $m\otimes m^{\prime }-m^{\prime }\otimes m$. This created commutative graded $R$-algebra $\mathcal{S}(M)$, called the symmetric algebra. We also introduced new notation, writing, for example, simply $m_1{m}_2$ for the coset $m_1\otimes m_2+\mathcal{C}(M)$.

We noted how to think about this construction in the vector space example.

Finally, we recalled a bit about the classical determinant function and started the construction of our final algebra, namely the exterior algebra. We didn't get far, however, before running out of time. We'll talk more about the exterior algebra next class.

Concepts

• Tensor algebras
• Symmetric algebras
• Exterior algebras

References

• Dummit & Foote: Section 11.5