# Defining the exterior algebra

To define a quotient algebra $\mathcal{A} / I$, with $\mathcal{A}$ an algebra over a field $K$, I thought that $I \subset A$ is a condition which had to be required. However, I read that the exterior algebra is the quotient algebra $\Lambda(V) = T(V) / I(V)$, where $T(V)$ is the tensor algebra, $$T(V) = \bigoplus_{k = 0}^{\infty} V^{\otimes k} \left(= K \oplus \bigoplus_{k = 1}^{\infty} \left(\bigotimes_{i = 1}^k V\right)\right)\mbox{,}$$ and $I(V)$ is the two-sided ideal generated by elements $v \otimes v$, with $v \in V$. Thus, the condition $I(V) \subset T(V)$ is not clear no me, as an element $v$ in $T(V)$ is of the form $(v_1 , \ldots , v_n , 0 , \ldots)$, where $v_j$ is $V^{\otimes j}$ for all $j = 1 , \ldots , n$ and (as my thought) $$I(V) = \left\{\sum_{i = 1}^n a_i \otimes (v_i \otimes v_i) \otimes b_i : n \in \mathbb{N}, a_i , b_i \in T(V) \mbox{ and } v_i \in V \mbox{ for all } i = 1 , \ldots , n\right\}\mbox{.}$$ Then I must be wrong in some of my statements because I cannot see any sense to the definition of $\Lambda(V)$. Can you help me? Thank you very much in advance.

Well, everything you've written is absolutely true.

It looks a bit confusing how you denote elements of $T(V)$ as finite sequences $(v_0, v_1, v_2, \dots, v_n, 0, 0, \dots)$, and elements of $I(V)$ as actual sums of decomposable tensors $\sum a_i \otimes v_i \otimes v_i \otimes b_i$, since these two are actually the same.

Let's refresh what direct sums are. For a sequence $V_i$ of modules (or vector spaces), their direct sum is denoted as $\bigoplus V_i$, and technically consists of sequences $(x_1, x_2, x_3, \dots)$, where $x_i \in V_i$. However, each $V_i$ gets embedded into the direct sum, and, abusing the notation, we can speak about elements of $V_i$ as sitting directly in $\bigoplus V_i$. This way, one also writes the aforementioned sequence as $\sum x_i$, meaning that the $x_i$'s are naturally embedded into $\bigoplus V_i$.

Now, an element of $T(V)$, the latter being the direct sum $\bigoplus V^{\otimes k}$, is a sequence $(x_0, x_1, x_2, \dots)$, but remembering the embedding above, one can also write $\sum x_i$. Here, $x_i \in V^{\otimes i}$.

Finally, consider an element of $I(V)$. It is also a sum of tensors. Specifically, if $a \in V^{\otimes i}$, $b \in V^{\otimes j}$, and $v \in V = V^{\otimes 1}$, then $a \otimes v \otimes v \otimes b \in V^{\otimes (i + j + 2)}$. Thus, the element $a \otimes v \otimes v \otimes b$ sits in the $(i+j+2)$-th position of a sequence as an element of $T(V)$. Note that in the sum $\sum a_i \otimes v_i \otimes v_i \otimes b_i$ there may be several elements of the same tensor rank, thus going to the same position in the sequence after they get summed (as elements of $V^{\otimes n}$).

If $a,b$ are not indecomposable, then decompose them into homogeneous parts: $a = \sum a_i$, $b = \sum b_j$, where, say, $a_i \in V^{\otimes i}$ and $b_j \in V^{\otimes j}$. Then, $a \otimes v \otimes v \otimes b = \sum\limits_i\sum\limits_j a_i \otimes v \otimes v \otimes b_j$, and $a_i \otimes v \otimes v \otimes b_j \in V^{\otimes(i+j+2)}$.

• In which way is ''each $V_i$ gets embedded into the direct sum'' true? Do you consider, for instance, the map $j_i : V_i \to \bigoplus_k V_k$, given by $j_i(v) = (0 , \ldots , 0 , \overset{(i)}{v} , 0 , \ldots)$? – joseabp91 Apr 26 '18 at 10:11
• @joseabp91 Yes, exactly this map. – lisyarus Apr 26 '18 at 11:25
• If $a = \sum_i a_i$ and $b = \sum_j b_j$ are in $T(V)$, the definition $a \otimes b$ is defined by $\sum_i \left(\sum_j (a_i \otimes b_j)\right)$ and it has sense because $a_i \otimes b_j$ is previously defined. But how can you define $a \otimes v$ or $v \otimes b$, with $v \in V$? – joseabp91 Apr 26 '18 at 11:44
• @joseabp91 $V$ is the same as $V^{\otimes 1}$ by definition. – lisyarus Apr 26 '18 at 12:07
• Yes but the elements in $T(V)$ are of the form $(v_1 , v_2 , \ldots)$, with $v_i \in V^{\otimes i}$. Thus, do you define $a \otimes v$ as $(v_1 \otimes v , v_2 , \ldots)$? – joseabp91 Apr 26 '18 at 12:13