Vertical Fourier decomposition for skew-Hermitian 1-forms

Question

In an arXiv preprint [2108.05125v1], the authors use the following vertical Fourier decomposition (page 7 therein). Let $(M,g)$ be a Riemannian surface and $SM$ be its unit tangent bundle. Denote by $V$ the vertical vector field on $SM$, i.e. $V = \partial/\partial \theta$ generated by the angle coordinate, $(x,\theta) \in SM$. The authors then write $L^2(SM,\mathbb{C}^n) = \bigoplus_{k \in \mathbb{Z}} H_k$ as an orthogonal decomposition into the eigenspaces $H_k$ of $-iV$ corresponding to eigenvalue $k$.

Now, suppose $A = \sum_{k \in \mathbb{Z}} A_k \in C^{\infty}(SM,\mathfrak{u}(n))$, where $\mathfrak{u}(n)$ is the space of skew-Hermitian matrices. The authors claim on page 31 that $A_{-k} = -A_k^*$ for all $k \in \mathbb{Z}$ due to the skew-Hermitian property. I cannot verify this equality. Let me give an example that illustrates my problem. Suppose $\omega$ is a $\mathfrak{u}(n)$-valued 1-form on $M$ and consider $A(x,v) = \omega_x(v)$. Then $A$ is linear in its second argument. As such, for any vector field of the form $Z(x,v) = (0,\lambda v)$ in the canonical splitting of $TSM$ into its horizontal and its vertical component, we have $Z(A)(x,v) = dA(Z)(x,v) = A_x(\lambda v) = \lambda A_x(v)$. In other words, $A$ is an eigenfunction of $Z$ for the eigenvalue $\lambda$. This eigenvalue depends only on $Z$.

So, regardless of whether $A$ is skew-Hermitian or not (we only used linearity of $A$ in $v$), I do not see how there could be 1-forms that are eigenfunctions of $V$ for the eigenvalue $1$ and also other 1-forms that are eigenfunctions of $V$ for the eigenvalue $-1$, which should be the case given $A_{-1} = -A_1^*$.

Related to this problem, I am not sure how to understand $-iV$: while $V$ is a vector field on $SM$, it seems that $-iV$ lives in $TTM$ but not in $TSM$.

Any help would be greatly appreciated.

Answer 1

If $SM$ is trivial (such that you can view it as $M\times \mathbb S^1$ and use the angle $\theta$ to describe the second variable), the $k$th Fourier mode of $\mathbb A\in C^\infty(SM,\mathbb C^{n\times n})$ is given by $$ \mathbb A_k(x)=\int_{0}^{2\pi}\mathbb A(x,\theta)e^{-ik\theta}d\theta. $$ Take the complex conjugate (which flips the sign in the exponential) and transpose to obtain $(\mathbb A_k)^*=(\mathbb A^*)_{-k}$. If $\mathbb A$ is skew Hermitian, this means that $\mathbb A_k^*=-\mathbb A_{-k}$.

I don't quite understand your computation with $Z$, especially the equality $dA(Z)(x,v)=A_x(\lambda v)$ does not seem to make sense. Keep in mind that up to this point it should work for any smooth function $\mathbb A = A$ on $SM$, such that $\mathbb A(x,\lambda v)$ is not even well-defined unless $\vert \lambda \vert=1$. Say the vector field $Z$ actually equals $V$, such that its flow is given by $\psi_t(x,v)=(x,e^{it}v)$. Then $$ V\mathbb A=\frac d {dt}\big\vert_{t=0} \psi_t^*\mathbb A=\frac d {dt}\big\vert_{t=0} \mathbb A(x,e^{it}v)=? $$ If $\mathbb A(x,v)=\omega_x(v)$ for a $1$-form that is complex linear in $v$, you can continue the computation and obtain $i\mathbb A$, which is to say that $\mathbb A = \mathbb A_1 \in H_1$. Being complex linear means that $\omega$ is a $(1,0)$-form (so it looks like $f dz$ in a local holomorphic chart of the Riemann surface $M$). More generally, the splitting $\Omega^1(M)=\Omega^{1,0}\oplus \Omega^{0,1}$ corresponds to the splitting $H^{1}\oplus H^{-1}$ for its lift to $SM$.

Regarding the meaning of $-iV$: We frequently view vector fields as first order differential operators, that is, $V\colon C^\infty(SM,\mathbb C)\rightarrow C^\infty(SM,\mathbb C)$ acts by taking the vertical derivative. In this viewpoint it is clear how to multiply by $i$. Your confusion might arise because $i$ is sometimes also used to denote a $90$ degree rotation in $T_xM$. If we rotate $V(x,v)$ inside the vertical subbundle in $T_{(x,v)}(TM)$ we obtain a vector field $V_\perp$, which indeed fails to be tangent to $SM$. But this vector field is of no relevance in the paper.