# If and how is this related to Grassmanians?

Given a $$n$$ dimensional vector space over the finite field $$F_q$$, called $$V(F_q)$$, and a set of $$M$$ vectors $$\vec c_m=(c_0,c_1,...c_{n-1})^T$$ that fulfill $$(\sum c_k )=0$$. Further the complete set of $$\{\vec c_m\}$$ is symmetric under all permutations of elements of a given $$\vec c_m$$.

I'm searching $$\vec s\in V(F_q)$$, the smallest (in the sense of the taxicab norm) element of $$V(F_q)$$ which has:

$$\vec s \cdot \vec c_m \neq 0 \bmod q, \forall m$$

If each $$\vec c_m$$ makes up an $$n-1$$ dimensional subspace, orthogonal to $$\vec c_m$$, my feeling says that $$\vec s$$ lies in the complement of union of all these subspaces. If and how is this related to Grassmanians? I ask because the application of Gaussian Binomial Coefficients tells us that

the Gaussian binomial coefficient $${n \choose k}_{q}}$$ counts the number of $$k$$-dimensional vector subspaces of an $$n$$-dimensional vector space over $$F_q$$ (a Grassmannian).

Is the union of all these subspaces a Grassmanian?

• "If each $\vec{c}_m$ makes up an $n−1$ dimensional subspace". What do you mean by this? The vector $\vec{c}_m$ belongs to an $n$-dimensional vector space, which you denote by $V(F_q)$. Which $n - 1$ dimensional subspace of $V(F_q)$ are you referring to? – Michael Albanese Jun 19 at 22:09
• @MichaelAlbanese I clarified: If each $\vec c_m$ makes up an $n-1$ dimensional subspace, orthogonal to $\vec c_m$, my feeling says that $\vec s$ lies in the complement of union of all these subspaces. – draks ... Jun 20 at 13:41
• I agree with that statement. – Michael Albanese Jun 20 at 19:10
• What makes you think that the problem you are trying to solve has anything to do with grassmannians? – Michael Albanese Jun 22 at 14:25
• I don't think that there is any reasonable way to frame this problem in terms of grassmannians. – Michael Albanese Jun 28 at 19:04