Embedding the exterior algebra into the tensor algebra

Question

In Griffiths & Harris Principles of Algebraic Geometry on page 28 it seems like $\alpha \otimes \beta - \beta \otimes \alpha = 2\alpha \wedge \beta$ (the equation where they express $\omega$ in terms of real differential forms). But from what I read in Wikipedia it should be $\alpha \otimes \beta - \beta \otimes \alpha = \alpha \wedge \beta$. Is this by mistake in G&H or is this just a different convention on how to embed $\Lambda^2(V)$ into $V^{\otimes 2}$.

Answer 1

The exterior power $\Lambda^2 V$ is defined as the quotient $$ \Lambda^2 V = V^{\otimes 2} \big/ \operatorname{span}\{ v \otimes v \mid v \in V \}.$$ As a quotient, there is a natural surjection $$ p: V^{\otimes 2} \twoheadrightarrow \Lambda^2 V, \quad p(v \otimes w) = v \wedge w.$$ There is also a natural injection in the opposite direction: $$ q: \Lambda^2 V \hookrightarrow V^{\otimes 2}, \quad q(v \wedge w) = v \otimes w - w \otimes v.$$ The image of $q$ is precisely the alternating tensors $\operatorname{Alt}^2(V) \subseteq V^{\otimes 2}$, which we can describe using the flip map $\sigma: V^{\otimes 2} \to V^{\otimes 2}$, $\sigma(v \otimes w) = w \otimes v$, in the following way: $$ \operatorname{Alt}^2(V) = \{\omega - \sigma(\omega) \mid \omega \in V^{\otimes 2}\} \subseteq V^{\otimes 2}.$$ If we compose the maps $pq: \Lambda^2 V \to \Lambda^2 V$, we get $$ v \wedge w \mapsto v \otimes w - w \otimes v \mapsto v \wedge w - w \wedge v = 2 v \wedge w,$$ and hence $pq$ is multiplication by 2. We can also see that the composition $qp$ is multiplication by $2$ on the elements of $\operatorname{Alt}^2(V)$.

Now we want to pick an isomorphism $\Lambda^2 V \cong \operatorname{Alt}^2(V)$. The map $q: \Lambda^2V \to \operatorname{Alt}^2(V)$ is usually the one that is chosen, since it works in all characteristics and it looks the nicest: $v \wedge w$ corresponds to $v \otimes w - w \otimes v$. In characteristic not 2, we have $p/2 = q^{-1}$, but in characteristic $2$ we cannot write down $q^{-1}$ in terms of $p$.

The other convention is to choose $p: \operatorname{Alt}^2(V) \to \Lambda^2 V$ as the isomorphism, which is indeed an isomorphism outside of characteristic 2. Using this choice, the element $v \wedge w$ corresponds to $\frac{1}{2}(v \otimes w - w \otimes v)$, and the inverse of $p$ is given by $p^{-1} = q/2$.