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Let $V$ be a vector space with dimension $n$ over a field $K$. The exterior algebra $\Lambda(V)$ of the vector space $V$ is the direct sum of the exterior powers $\Lambda^k(V),\quad k\in\overline{0,n}$. Then an element $x\in\Lambda(V)$ has the form $(x_0,\dots,x_n)$, where $x_i \in \Lambda^i(V)$ is the $i$-th homogeneous component of $x$. As each exterior power is itself a vector space over $K$, so is their direct sum with the point-wise addition and multiplication with a scalar.

How does the exterior product carry over the direct sum? What does $(x_0,\dots,x_n)\wedge (y_0,\dots,y_n)$ stand for?

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  • $\begingroup$ Yes, I know that. But if $x\in \Lambda(V)$, then $x=(x_0,\dots,x_n)$, where $x_i\in\Lambda^i(V)$. Really my question is how the product in algebras carries over to the direct sum of those algebras. Is it defined componentwise? $\endgroup$ – superAnnoyingUser May 7 '14 at 10:23
  • $\begingroup$ It is defined component-wise. The "right" way to think of your element is as $\sum_j x_j$. Then you define $\wedge$ so as to make it distributive and take $\Lambda^j(V) \times \Lambda^k(V) \to \Lambda^{j+k}(V)$. $\endgroup$ – jdc May 7 '14 at 12:43
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Since $\Lambda(V) = \Lambda^0(V) \oplus \cdots \oplus \Lambda^n(V)$, a general element $x \in \Lambda(V)$ looks like $x = x_0 + \cdots + x_n$, where $x_k \in \Lambda^k(V)$ for $k \in \{0, \dots, n\}$. The product $\wedge$ on $\Lambda(V)$ is the image of $\otimes$ on $T(V)$ (the tensor product on the tensor algebra) when its quotient is taken by the $2$-sided ideal of "repeating tensors", generated by tensors of the form $(v \otimes v)$ where $v \in V$. Since $\otimes$ acts distributively, (not component-wise), so does $\wedge$: $$ (x_0 + \cdots + x_n) \wedge (y_0 + \cdots + y_n) = \sum_{k=0}^n x_k \wedge (y_0 + \cdots + y_n) = \sum_{k=0}^n \sum_{j=0}^n x_k \wedge y_j. $$ This looks just like the case in which $\wedge$ is replaced by $\otimes$ and the $x_k$ and $y_j$ are homogeneous tensors of equal degree. As with polynomials, we like to collect terms with equal degree, which can be done via $$ = \sum_{k=0}^{n} x_k \wedge y_{n-k}. $$

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The product is defined component-wise, i.e. $$(x_0,\dots,x_n)\wedge (y_0,\dots,y_n) = (z_{0}, \ldots z_{n})$$ where $$z_{i} = \sum_{j = 0}^{i} x_{j} \wedge y_{i-j}$$

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  • $\begingroup$ The formula is correct, but I wouldn't call it "component-wise"! $\endgroup$ – darij grinberg Aug 1 '16 at 19:05

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