0
$\begingroup$

I am trying to find the conditions under which a complex circle manifold intersects with a linear subspace. I have a linear subspace of dimension T-K (defined as the null subspace of a $K \times T$ complex matrix) (i.e., a subspace of the space of complex vectors of size $T$, $\mathbb{C}^T$). I have control over $T$ but $K$ is fixed. The complex circle manifold is defined as $\mathcal{C} =\{\mathbf{x} \in \mathbb{C}^T|\; |x_1| = |x_2|=\cdots = |x_T|=1\}$. It has a dimension of $T/2$ as far as I see it (complex dimensions). Both the subspace and the complex circle manifolds are submanifolds of the Euclidean space. I am confused with the concept of dimensions counting (i.e., saying that $(T-K) + (T/2) > T$ would guarantee that the intersection will be non-empty) and I am not sure if it fits here. The reason for my confusion is that if I count the dimensions of two hyperspheres of different radii they will, of course, exceed $T$ dimensions, yet obviously, there is no intersection between them.

Please accept my apology as I am not a math expert and I might be missing something clear.

$\endgroup$
7
  • 1
    $\begingroup$ Your "complex circle manifold" is a torus: $\mathcal{C} = S^1 \times \cdots \times S^1 = \mathbb{T}^T$ of real dimension $T$. However, $\mathcal{C}$ is generally not a complex manifold, so it's not really correct to say that $\mathcal{C}$ has "complex dimension $T/2$." (To be more technical, your $\mathcal{C}$ is a Lagrangian submanifold of $\mathbb{C}^T$, but not a complex submanifold of $\mathbb{C}^T$.) Separately, your question about the dimension of intersecting submanifolds has to do with "transversality." $\endgroup$ Commented Dec 30, 2020 at 16:38
  • $\begingroup$ Thanks a lot for your responses. I really appreciate your help. What I really care about is whether the subspace and the complex circle would intersect and under what conditions in terms of T and K? I am not sure if this doable through dimensions counting or I should seek some other approach to prove that the intersection is non-empty $\endgroup$
    – kseddik
    Commented Dec 30, 2020 at 16:48
  • $\begingroup$ @JesseMadnick I did read about transversality but as far as I see it, it relates to some structures at the points of intersection. I am looking for conditions to guarantee that there is a non-empty intersection. $\endgroup$
    – kseddik
    Commented Dec 30, 2020 at 17:03
  • $\begingroup$ @JesseMadnick I have modified the problem description. The subspace in my problem is defined as the null space of a complex matrix of dimension $K\times T$ (the matrix defines a set of $K$ linear constraints). The solution should be in the null space of this matrix and at the same time, it should be such that each element in the vector has a unit norm. This is why I need to get the condition(s) under which the intersection is non-empty controlling $T$ (I just need to prove that a solution exists). $\endgroup$
    – kseddik
    Commented Dec 30, 2020 at 19:20
  • $\begingroup$ Your complex $K \times T$ matrix defines $K$ independent linear constraints if and only if its rank is $K$ (which requires $K \leq T$). In general, however, if your matrix has rank $< K$, then your matrix will define fewer than $K$ constraints (consider, e.g., the zero matrix). $\endgroup$ Commented Dec 30, 2020 at 19:45

1 Answer 1

0
$\begingroup$

Suppose $K = 1$. Consider the $1 \times T$ matrix $$A = \begin{bmatrix} 0 & 0 & \cdots & 0 & 1 \end{bmatrix}\!.$$ The nullspace of $A$ is the linear subspace $$L = \{\mathbf{z} = (z_1, \ldots, z_T) \in \mathbb{C}^T \colon z_T = 0\}.$$ Note that $L$ has complex dimension $T - 1$. In other words, $L$ has the largest possible complex dimension inside of $\mathbb{C}^T$ (without being $\mathbb{C}^T$ itself). Nevertheless, $L$ does not intersect the torus $$\mathcal{C} = \{\mathbf{z} \in \mathbb{C}^T \colon |z_1| = 1, \ldots, |z_T| = 1\},$$ because no point of $L$ has the property that $|z_T| = 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .