4
$\begingroup$

Background

The tensor space of type $(r,s)$ associated with $V$ is the vector space $$\underbrace{V\otimes \ldots \otimes V}_{\text{r copies}} \otimes \underbrace{V^* \otimes \ldots \otimes V^*}_{\text{s copies}}.$$

The tensor algebra of $V$ is the direct sum $T(V)=\sum_{r,s} V_{r,s}$ for $r,s \geq 0.$

Let $C(V)$ be the subalgebra $\sum_{k=0}^{\infty} V_{k,0}$ of $T(V)$. Let $I(V)$ denote the two sided ideal in $C(V)$ generated by the set of elements of the form $v \otimes v,$ where $v \in V$ and set $$I_k(V)=I(V) \cap V_{k,0}.$$

The exterior algebra $\Lambda(V)$ is the graded algebra $C(V)/I(V)$. If we set $$\Lambda_k(V)=V_{k,0}/I_k(V)$$ for $k \geq 2$ and $\Lambda_0(V)=\mathbb{R}, \ \Lambda_1(V)=V$ then $$\Lambda(V)=\sum_{k=0}^{\infty} \Lambda_k(V).$$

Denote multiplication in the algebra $\Lambda(V)$ by $\wedge$

Questions

I know very little abstract algebra so a lot of this construction is strange to me. I have the following questions:

  1. What is the point of doing the intersection $I(V) \cap V_{k,0}$ when defining $I_k(V)$?

  2. Why is the exterior algebra graded? Here is what I attempted to do:

    Let $u \in \Lambda_k(V)$ and $v \in \Lambda_l(V)$ We should show that $u \wedge v \in \Lambda_{k+l}.$ I think $u \in \Lambda_k$ means that $u$ is the equivalence class $u+I_{k}(V)$ and similarly $v$ is the equivalence class $v+I_{l}(V).$ I also think that $u \wedge v$ corresponds to the equivalence class of $u \otimes v$ in (I don't know what space).

As you can see, my understanding of this construction is very weak. I think if I see $2$ clearly I will understand how to work with the exterior algebra. To keep this question unambiguous, please consider my attempt with question $2$ above and try to explain, explicitly, what an element in $\Lambda_n(V)$ looks like (and how I can see this from the construction above) and also to explain what $u \wedge v$ means in terms of the quotient construction.

$\endgroup$
  • 1
    $\begingroup$ 1. You can define the algebra $\Lambda\left(V\right)$ straight away as the quotient $T\left(V\right) / I\left(V\right)$; then you don't need to define $I_k\left(V\right)$. But this way you get $\Lambda\left(V\right)$ merely as an algebra, not as a graded algebra. You can still define the grading as an afterthought, but the author of the approach wanted to get it right at the beginning, whence the introduction of $I_k\left(V\right)$. $\endgroup$ – darij grinberg Apr 3 '16 at 1:54
  • 1
    $\begingroup$ ... (in fact, this difference is a linear combination of pure tensors with two equal adjacent tensorands). Hence, the projections of $x \otimes y \otimes z$ and $-y \otimes x \otimes z$ on $\wedge^3\left(V\right)$ are equal. I have chosen two pure tensors to keep this example simple, but of course not every tensor is pure, so there are more complicated examples. $\endgroup$ – darij grinberg Apr 3 '16 at 1:59
  • 1
    $\begingroup$ If $u \in V^{\otimes k}$ and $v \in V^{\otimes l}$, then $u \otimes v \in V^{\otimes \left(k+l\right)}$. This is not a particular case of your construction with $u+N$ and $v+N$, because here $u$ and $v$ are understood to live in two different spaces. $\endgroup$ – darij grinberg Apr 3 '16 at 2:33
  • 1
    $\begingroup$ One thing that might be confusing you: An element of $\wedge^k V$ can be viewed either as an element of $V^{\otimes k} / I_k\left(V\right)$, or as an element of the bigger space $\wedge V = \Lambda\left(V\right) = T\left(V\right) / I\left(V\right)$. In the former case, it is a congruence class of an element of $V^{\otimes k}$ modulo $I_k\left(V\right)$; in the latter case, it is a congruence class of an element of $T\left(V\right)$ modulo $I\left(V\right)$. When you regard $u$ and $v$ in the latter fashion (i.e., as congruence classes of elements of $T\left(V\right)$), then ... $\endgroup$ – darij grinberg Apr 3 '16 at 2:36
  • 1
    $\begingroup$ ... you indeed can multiply them by the rule $\left(u+N\right)\left(v+N\right) = uv+N$ (notice that $N$ here is $I\left(V\right)$). When you regard $u$ and $v$ in the former fashion, however, this pattern does not apply, since it would involve three different $N$'s. Instead, what you do is define their product by $\left(u+I_k\left(V\right)\right) \left(v + I_l\left(V\right)\right) = u \otimes v + I_{k+l}\left(V\right)$ (please don't put a tensor sign on the left hand side). This can also be seen as a special case of a general rule, but it's not as simple as "algebra modulo ideal". $\endgroup$ – darij grinberg Apr 3 '16 at 2:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.