I just want to check that my understanding is correct. Suppose you have a vector space. Suppose we have a linear operator $J : V \rightarrow V$ such that $J^2 = -1$. Then V as a complex vector space admits the following basis

$$\{e_1,J \circ e_1,\ldots,e_n,J \circ e_n\}$$

Consider the dual basis given by $$\{\alpha_1, \alpha_1 \circ J,\ldots,\alpha_n,\alpha_n \circ J\}$$

The following when applied to the basis element will have value = 2. That is:

$$(\alpha_1 \wedge \alpha_1 \circ J \wedge \ldots \wedge \alpha_n \circ J)(e_1,J \circ e_1,\ldots,e_n,J \circ e_n) = 2 > 0$$

Thus, it has a natural orientation.

  • $\begingroup$ Your linear operator $J$ should be an morphism of vector spaces? Moreover, how can we assume that $V$ is a complex vector space? $\endgroup$ – Jonas Lenz Nov 25 '17 at 16:07
  • $\begingroup$ @JonasLenz Given a (real) vector space $V$ as in the question, a complex structure $J$ on $V$ realizes $V$ as a complex vector space by extending the real scalar multiplication map by declaring $i \cdot v := Jv$ and using linearity. $\endgroup$ – Travis Nov 25 '17 at 16:20

The mechanical idea of the proof (namely, forming the top form $e_1 \wedge J e_1 \wedge \cdots \wedge e_n \wedge J e_n$ and declaring it to be positive) is certainly a reasonable way to carry this out, but there are a few issues with the argument as written.

  1. If $\dim_{\Bbb R} V = 2n$, then the complex structure $J : V \to V$ realizes $V$ as a complex vector space with dimension $\dim_{\Bbb C} V = n$; for emphasis, I'll write this complex vector space as $\widetilde V$, even though typically we usually use the same symbol for both spaces. In particular, a basis of $\widetilde V$ should have $n$ elements: $(e_1, \ldots, e_n)$. Given any such basis, one gets without making any additional choices a basis $(e_1, J e_1, \ldots, e_n, J e_n)$ of $V$, and it's this real basis that you're using to construct an orientation on $V$.

  2. As given here, the construction involves making a choice $(e_a)$ of (complex) basis of $\widetilde V$. To show that the orientation is natural, you need to show that the orientation you produce is independent of the basis you choose. (After all, fixing a basis $(f_a)$ of any vector space $W$ determines an orientation of $W$ by declaring $f_1 \wedge \cdots \wedge f_n$ to be positive, but if $\dim W > 0$ there are other choices basis that lead to the opposite orientation---indeed, in general vector spaces don't carry natural orientations.)

  3. You probably don't really need this fact for this argument anyway, but the cobasis $(\alpha_1, \alpha_1 \circ J, \ldots, \alpha_n, \alpha_n \circ J)$ you mention is not dual to the basis $(e_1, J e_1, \ldots, e_n, J e_n)$. Comparing the second elements, for example, gives $$(\alpha_1 \circ J) (J e_1) = \alpha_1(J^2 e_1) = \alpha_1(- e_1) = -\alpha_1(e_1) = -1 ,$$ not $1$.


Your Answer

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