Natural orientation of an underlying real vector space

Question

Consider the following argument from a book (included as an image to show exactly what is printed):

There are a number of minor typos here, but I'm mainly concerned with the computation of $\Delta(a_1,\ldots,a_n,ia_1,\ldots,ia_n)$. In the definition of $\Delta$ I'm assuming that the author is using the Grassmann product defined earlier in the book, which here would give $$\Delta(x_1,\ldots,x_{n+n})=\frac{(-i)^n}{(n!)^2}\sum_{\sigma\in S_{n+n}}\varepsilon_{\sigma}\Delta_E(x_{\sigma(1)},\ldots,x_{\sigma(n)})\overline{\Delta}_E(x_{\sigma(n+1)},\ldots,x_{\sigma(n+n)})\tag{1}$$ If I take $E=\mathbb{C}$ (so $n=1$), then from (1) I obtain $$\Delta(a_1,ia_1)=-2|\Delta_E(a_1)|^2<0$$ Even using the expression on the right-hand side of the first equality in the computation from the book, I obtain $$\Delta(a_1,ia_1)=-|\Delta_E(a_1)|^2<0$$ If I am correct, does someone know what the definition of $\Delta$ should be to yield the desired result (for all $n$)?

Answer 1

$ \newcommand\Ext{{\bigwedge}} \newcommand\R{\mathbb R} \newcommand\C{\mathbb C} \newcommand\conj[1]{\overline{#1}} \newcommand\form[1]{\langle#1\rangle} \DeclareMathOperator\linspan{span} \renewcommand\Re{\mathop{\mathrm{Re}}} \newcommand\tensor\otimes \newcommand\gtensor{\mathbin{\hat\tensor}} $

I think that $\Delta = (-i)^n\Delta_E\wedge\bar\Delta_E$ is a typo and should be $$ \Delta = {\color{red}{\frac1{(-i)^n}}}\Delta_E\wedge\bar\Delta_E = i^n\Delta_E\wedge\bar\Delta_E. $$ This is somewhat natural since (as Greub is trying to convince us) $$ \Delta_E(ia_1, \dotsc, ia_n)\wedge\bar\Delta_E(ia_1, \dotsc, ia_n) = i^n\Delta_E(a_1,\dotsc,a_n)\wedge\bar\Delta_E(ia_1,\dotsc,ia_n) $$ is non-zero and real. It could also be the case that Greub meant to write $$ \Delta(ia_1, ia_2, \dotsc, ia_n, a_1, a_2, \dotsc, a_n) $$ in which the case the definition of $\Delta$ as written by Greub is the one required.

In either case, the argument proceeds by suggesting to the reader that $\sum_{S_{2n}} = \sum_{S_n\times S_n}$ when evaluating this particular wedge product. This doesn't seem outlandish to me, but I also don't immediately see how that works.

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What I will do is give a more abstract approach, which completely avoids the tedious combinatorics. It would be a good exercise to pinpoint how this argument lines up with Greub's.

Let us consider the exterior algebra.

Choose non-zero $J \in \Ext^n E$. This induces a linear isomorphism $\cdot/J : \Ext^n E \cong \C$ via $$ J' = (J'/J)J,\quad J' \in \Ext^n E. $$ The notation $J'/J$ is merely meant to be suggestive. Such an isomorphism is a determinant function, and is equivalent to a choice of $J$ by choosing the unique $J'$ such that $J'/J = 1$. We will show that (given a notion of conjugation) $J$ naturally gives rise to an $I \in \Ext^{2n}E_\R$ by constructing a natural determinant function $\Delta \in \Ext^{2n}E_\R \to \R$, and we will also show that every such $I$ has the same orientation.

A conjugation on $E$ is not inherent to $E$ being a complex vector space; it is an additional structure. Once we have a conjugation map, there is a special subset of $E$ $$ \Re E := \{v + \conj v \;:\; v \in E\} $$ whose elements we call the real elements of $E$. This is not a subspace of $E$, but clearly is a subspace of $E_\R$. By $\Ext\Re(E)$ we mean the subspace of $\Ext E_\R$ generated by $\Re(E)$. It is evident that $$ E_\R = \Re E \oplus i\Re E, $$ and it follows that $\Ext E = \Ext\Re(E)\gtensor\Ext i\Re(E)$ where $\gtensor$ is the graded tensor product. In particular $$ \Ext^{2n} E_\R = \Ext^n\Re E\gtensor\Ext^n i\Re E. $$ The graded aspect of $\gtensor$ isn't important here; when considering just linear maps, $\gtensor$ is just like the usual tensor product. By this fact, we may construct a linear map $\Ext^n E_\R \to \R$ by specifying its values on pairs $(H, H')$ for $H \in \Ext^n\Re(E)$ and $H' \in \Ext^ni\Re(E)$ and ensuring bilinearity.

By the universal property of $\Ext E_\R$, the $\R$-linear isomorphism $\phi : E_\R \to E \subseteq \Ext E$ extends uniquely to an $\R$-algebra homomorphism $\Ext E_\R \to \Ext E$. Our candidate for $\Delta$ is then $$ H\tensor H' \mapsto \conj{\frac{\phi(H)}J}\frac{\phi(H')}J. $$ Here, the conjugation is meant to apply to the entire expression $\phi(H)/J$. Consider though the particular case where $H' = m_i(H)$ for $m_i : \Ext E_\R \to \Ext E_\R$ the outermorphism which multiplies vectors by $i$. Then $\phi(H') = i^n\phi(H)$ and $$ \conj{\frac{\phi(H)}J}\frac{\phi(H')}J = i^n\left|\frac{\phi(H)}J\right|^2. $$ So we divide by $i^n$. Define $\Delta : \Ext^{2n}E_\R \to \R$ by $$ \Delta(H\tensor H') = (-i)^n\conj{\frac{\phi(H)}J}\frac{\phi(H')}J $$ This still gives us a real number for any $H'$ since $H' = xm_i(H)$ for some $x \in \R$ by virtue of being an element of $\Ext^ni\Re(E) = m_i\left(\Ext^n\Re(E)\right)$. Observe now that for any non-zero $a \in \C$ $$ J'/(aJ) = \frac{J'/J}a, $$ hence if $\Delta'$ is the $E_\R$ determinant function associated to $aJ$ $$ \Delta'(H\tensor H') = (-i)^n\conj{\frac{\phi(H)}{aJ}}\frac{\phi(H')}{aJ} = \frac{(-i)^n}{|a|^2}\conj{\frac{\phi(H)}J}\frac{\phi(H')}J = \frac1{|a|^2}\Delta(H\tensor H'). $$ Since $1/|a|^2$ is positive, $\Delta'$ has the same orientation as $\Delta$. Thus, as promised, every $J$ induces the same orientation on $E_\R$.

We could also consider the alternative $\Delta$ $$ \Delta(H\tensor H') = i^n\frac{\phi(H)}J\conj{\frac{\phi(H')}J} $$ but this is in fact equivalent to the original since there is some $x \in \R$ such that $H' = xm_i(H)$ and $$ i^n\frac{\phi(H)}J\conj{\frac{\phi(H')}J} = \frac{\phi(m_i(H))}Jx(-i)^n\conj{\frac{\phi(H)}J} = (-i)^n\frac{\phi(H')}J\conj{\frac{\phi(H)}J}. $$

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It is interesting to note that the same argument goes through if we replace every instance of $i$ with any $\xi \in \C\setminus\R$. Whether every choice of $\xi$ induces the same $\Delta$ as $i$ does is a question I leave alone for now.

Answer 2

In light of part of Nicholas' answer, and Ted's comment, I think Greub should have defined $\Delta$ by $$\Delta=\frac{i^n}{2^n}\Delta_E\wedge\overline{\Delta}_E$$ Then equation (1) in my question becomes $$\Delta(x_1,\ldots,x_{n+n})=\frac{i^n}{2^n(n!)^2}\sum_{\rho\in S_{n+n}}\varepsilon_{\rho}\Delta_E(x_{\rho(1)},\ldots,x_{\rho(n)})\overline{\Delta}_E(x_{\rho(n+1)},\ldots,x_{\rho(n+n)})\tag{1'}$$ Taking $x_k=a_k$ and $x_{n+k}=ia_k$ for $1\le k\le n$, what are the non-zero terms in the sum on the right side of (1')? First observe that in such a term the subscripts associated with the $a$'s and $ia$'s (not the $x$'s) occurring within $\Delta_E$ must be distinct, lest there be a linear dependence relation over $\mathbb{C}$ which causes $\Delta_E$, and hence the term, to be zero. Similarly for $\overline{\Delta}_E$. If the subscripts within each of $\Delta_E$ and $\overline{\Delta}_E$ are strictly increasing (call this an increasing term), then the term is nonzero and for each $1\le k\le n$ we must have either $$\rho(k)=k\quad\text{and}\quad\rho(n+k)=n+k\tag{2}$$ or $$\rho(k)=n+k\quad\text{and}\quad\rho(n+k)=k\tag{3}$$ Note there are $2^n$ such permutations $\rho$, and $\varepsilon_{\rho}=(-1)^p$ where $p$ is the number of $k$ for which (3) holds, which is just the number of $i$'s appearing within $\Delta_E$. These $i$'s can be moved over to $\overline{\Delta}_E$ by multiplying by $(-1)^p=\varepsilon_{\rho}$, since $\Delta_E$ is $\mathbb{C}$-multilinear and $\overline{\Delta}_E$ is conjugate $\mathbb{C}$-multilinear. Now every non-zero term is obtained from a unique increasing term by applying unique signed permutations $\sigma,\tau\in S_n$ to $\Delta_E$ and $\overline{\Delta}_E$, respectively. So we obtain $$\begin{align*} \Delta(a_1,\ldots,a_n,ia_1,\ldots,ia_n)&=\frac{i^n}{2^n(n!)^2}\cdot2^n\cdot\sum_{\sigma,\tau\in S_n}\varepsilon_{\sigma}\varepsilon_{\tau}\Delta_E(a_{\sigma(1)},\ldots,a_{\sigma(n)})\overline{\Delta}_E(ia_{\tau(1)},\ldots,ia_{\tau(n)})\\ &=\frac{i^n}{(n!)^2}\cdot(n!)^2\cdot\Delta_E(a_1,\ldots,a_n)\overline{\Delta}_E(ia_1,\ldots,ia_n)\\ &=\Delta_E(a_1,\ldots,a_n)\overline{\Delta}_E(a_1,\ldots,a_n)\\ &=|\Delta_E(a_1,\ldots,a_n)|^2>0 \end{align*}$$ A bit tedious, not as slick as an abstract argument, but it seems to work.