I read up on the following blog, M. Gillespie's paper and the Hatcher's textbook(K-theory and vector bundle) because why a Grassmannian is a CW-complex. I have some questions about this process. But like a title my question is only restricted to $0$-skeleton (later I will upload subsequent questions about what is an attaching map and how to make $X^2$ strcture and so on...). Also I will focus on $\operatorname{Gr}(4,2)$
- what is a n-cell structure : To begin with, anyway , if a Grassmannian is a really CW-complex, clearly the $n$-disk $D_{\alpha}^n$ exists. According to the M. Gillespie's paper, an open $n$-disk cell is correspondent to the Schubert cell, In order to explain what is a Schubert cell, the author explains a partition, $\lambda:=(\lambda_1,....,\lambda_n) $ a nonincreasing sequence of nonnegative integers
Then next, the author explains a Schubert cell, denoted by $\Omega_{\lambda}^{o}$, the set of points of $\operatorname{Gr}(n,k)$ whose row echelon matrix has corresponding partition $\lambda$. Next, the author calls the closure of Schubert cell , $\overline{\Omega_{\lambda}^{o}}:=\Omega_{\lambda} $, a Schubert variety. Schubert variety is likely correspondent to the closed ball. Frankly speaking, I am not sure why a Schubert cell analogously plays a role in an open $n$-disk. but I just believe the correspondence. Furthermore, the author explicitly express which element is a Schubert cell and Schubert variety.
Note.
$\Omega_{\lambda}^{o}= \left\{ V \in \operatorname{Gr}(n,k): \operatorname{dim}(V \cap <e_1,...e_r>)=i, for~ n-k+1-\lambda_{i} \le r \le n-k+1-\lambda_{i+1} \right\}$
$$\Omega_{\lambda} = \left\{ V \in \operatorname{Gr}(n,k): > \operatorname{dim}(V \cap <e_1,...e_{n-k+i-\lambda_{i}}>)\ge i \right\}(= \overline{\Omega_{\lambda}^{o}}) $$ where $e_1=(0,...,0,1), e_2=(0,...,0,1,0)$
Constructing $X^{0}$ skeleton : Then what is a $0$-skeleton? As I said, I want to construct $ \operatorname{Gr}(4,2)$ and Example 4.9 on M. Gillespie's paper introduced why $ \operatorname{Gr}(4,2)$ is a CW-complex structure. I cannot imagine which element should become a $0$-skeleton, $X^0$, but the author referred that $\Omega_{((n-k)^k)}$ is a $0$-skeleton. Since in our case, $n=4,k=2$. Thus, $\Omega_{(4)}$ is really a $0$-skeleton. (As I said, I can depend only on the Note because I cannot imagine what is $0$-skeleton) In that case, a partition is a single component $\lambda=(4)$ and $\lambda_1 =4$. However,
$\Omega_{(4)}= \left\{ V \in \operatorname{Gr}(4,2): \operatorname{dim}(V \cap <e_1,...e_{4-2+1-4}>)\ge 1 \right\}$ ($n=4,k=2,i=1, \lambda_{1}=4)$ $= \left\{ V \in \operatorname{Gr}(4,2): \operatorname{dim}(V \cap <e_1,...,e_{-1}>)\ge 1 \right\}$ [!]
$\Omega_{(4)}$ is clearly not well-defined. How can we define $e_{-1}$? Or is there any good suggestion that I understand what is a $0$-skeleton of $\operatorname{Gr}(4,2)$? ,
- dimension of $X^0$ : Meanwhile, what about the dimension of $0$-skeleton? clearly and naturally, the dimension should be $0$. In fact, on the given paper, there are following information about the dimension (Remark 3.8 on the page $9$)
$(1)$ $\operatorname{dim}(\Omega_{\lambda}^{o})=\operatorname{dim}(\operatorname{Gr}(n,k))-|\lambda|=k(n-k)-|\lambda| $
(where $|\lambda|=\sum_{i} \lambda _i $, and we call $|\lambda|$ a size of partition $\lambda$)$(2)$ $\operatorname{dim}(\Omega_{\lambda}^{o})= \operatorname{dim}(\Omega_{\lambda} ) $
According to the paper, the dimension is based on a complex number, $\mathbb{C}$, because the author defines Grassmannian on the $k$-dimensional subspaces of $\mathbb{C}^n$(Def 3.1 on the page $7$) hence, more precisely, the above dimension is written as $\operatorname{dim}_{\mathbb{C}}(\Omega_{\lambda}^{o})=k(n-k)-|\lambda| $ but I will omit the field $\mathbb{C}$.
Based on the above box on the dimension and about my interesting Grassmannian, $\operatorname{Gr}(4,2)$, $$\operatorname{dim}(\Omega_{\lambda} )=2(4-2)-4=0$$
But, I think that it seems a very lucky case ; when considering general case, in other words, for any $n,k (k\le n)$ $\Omega_{((n-k)^k)}$ is a $0$-skeleton. In that case, $|\lambda|=(n-k)^k$. Then, in that case,
$$\operatorname{dim}(\Omega_{\lambda} )=k(n-k)-(n-k)^k $$
but, the dimension generally would not be zero.... I cast doubt whether $\Omega_{((n-k)^k)}$ is a $0$-skeleton.
