Is there relation between Grassmann Manifold and Grassmann Algebra? I'm an EE student and I'm beginning to learn about the Grassmann Manifold.
As is known that the Grassmann Manifold is a space treating each linear subspace with a specific dimension in the vector space $V$ as a single point, for example we can represent the set of all $k$-dimensional linear subspaces $X$ in the n-dimensional vector space $V$ as Grassmann Manifold $Gr(k,n)$, and treat each $X\in Gr(k,n)$ as a point in this local Euclidean space.
When I'm exploring materials about Grassmann Manifold, lots of discussions about the so-called Grassmann Algebra just come to me. I learned that Grassmann Algebra deals with a so-called Exterior Product and is difined on a space that is different from the common vector spaces.
So I'm wondering about the relationship between Grassmann Manifold and Grassmann Algebra? Is there any algebra defined for the space of Grassmann Manifold?
 A: Just for the sake of completeness the definition of Grassmann algebra is recalled below and then the geometric interpretation of wedge operator is presented (which reveals the relation between Grassmann algebra and Grassmann manifold) 
Assume $\textbf{V}$ is an n-dimensional vector space over the field $\mathbb{R}$. Suppose that the space of all covariant tensors of order $r$ is denoted by $\mathcal{T}^{r}(\mathbf{V})$. Then the space $(\mathcal{T}(\mathbf{V}) , + , \otimes )$ forms an $\mathbb{R}$-algebra, where 
\begin{equation}
\mathcal{T}(\mathbf{V})=\mathcal{T}^{0}(\mathbf{V})\bigoplus \mathcal{T}^{1}(\mathbf{V})\bigoplus \mathcal{T}^{2}(\mathbf{V})\bigoplus ...     
\end{equation}
For each $r>0$ define ${\bigwedge}^{r}\mathbf{V} \subset \mathcal{T}^{r}(\mathbf{V})$ as the set  of alternating covariant tensors of order $r$. By taking the direct sum over ${\bigwedge}^{r}\mathbf{V}$ for all possible $r$, one achieves the following space:
\begin{equation}
{\bigwedge}\mathbf{V}={\bigwedge}^{0}\mathbf{V}\bigoplus {\bigwedge}^{1}\mathbf{V}\bigoplus {\bigwedge}^{2}\mathbf{V}\bigoplus\ldots\bigoplus\bigwedge^{n}\mathbf{V}  
\end{equation}
Under the addition and scalar product ${\bigwedge}\mathbf{V}$ is a vector subspace of $\mathcal{T}^{r}(\mathbf{V})$, and by defining the following product: 
\begin{eqnarray}
\wedge : {\bigwedge}^{\mathrm{r}}\mathbf{V} \times {\bigwedge}^{\mathrm{s}}\mathbf{V} \longrightarrow {\bigwedge}^{\mathrm{r+s}}\mathbf{V} \end{eqnarray}
\begin{eqnarray*}
(\phi, \psi)\longmapsto \frac{(\mathrm{r+s})!}{\mathrm{r}!\mathrm{s}!}\mathcal{A}(\phi\otimes\psi)
\end{eqnarray*}
where, $\mathcal{A}$ is defined as a linear transformation on $\mathcal{T}^{r}(\mathbf{V})$ 
\begin{eqnarray}
(\mathcal{A}\Phi)(\mathbf{v}_{1}, \mathbf{v}_{2}, ... , \mathbf{v}_{n})=\frac{1}{r!}\sum \mathrm{sgn}(\sigma)\Phi(\mathbf{v}_{\sigma(1)},..., \mathbf{v}_{\sigma(1)})
\end{eqnarray} 
the space $(\bigwedge(\mathbf{V}) , + , \wedge )$ is called the Grassmann algebra or exterior algebra.
Now here is the geometrical interpretation of the wedge product (which somehow shows the relation between the Grassmann algebra and Grassmann geometry):
Since, assigning geometrical objects to the algebraic expressions should be done uniquely the following map is desired to be one to one:
\begin{eqnarray}
\phi : G(r,n)\longrightarrow\mathbb{P}({\bigwedge} ^{\mathrm{r}}\mathbb{K}^{n})
\label{embedding}
\end{eqnarray} 
\begin{equation*}
\mathrm{span}(\mathbf{v_{1}}, ... , \mathbf{v_{r}})\longmapsto \mathbf{v_{1}}\wedge ... \wedge \mathbf{v_{r}}
\end{equation*}
It turns out that this map is an embedding of the Grassmann manifold into $\left(\begin{array}{c} n\\ r \end{array}\right)$-dimensional projective space $\mathbb{P}({\bigwedge} ^{\mathrm{r}}\mathbb{K}^{n})$. 
Suppose $\mathit{A}=\mathbf{a_{1}}\wedge ... \wedge\mathbf{a_{k}}$, then the vector space $\mathrm{span}(\mathbf{a_{1}}, ... , \mathbf{a_{k}})$ is called the support of $\mathit{A}$ and is denoted by $\bar{\mathit{A}}$.
Now, by considering $\mathit{A}=\mathbf{a_{1}}\wedge ... \wedge\mathbf{a_{k}}$ and $\mathit{B}=\mathbf{b_{1}}\wedge ... \wedge\mathbf{b_{j}}$ as two extensors of steps $k$ and $j$, the join of $\mathit{A}$ and $\mathit{B}$ is defined as the $(j+k)$-extensor
\begin{equation}
 \mathit{A}\wedge\mathit{B}=\mathbf{a_{1}}\wedge ... \wedge\mathbf{a_{k}}\wedge\mathbf{b_{1}}\wedge ... \wedge\mathbf{b_{j}}
\end{equation}
If the vectors $\{\mathbf{a_{1}}, ... ,\mathbf{a_{k}},\mathbf{b_{1}}, ... ,\mathbf{b_{j}}\}$ are linearly independent then
\begin{equation}
\bar{\mathit{X}}=\bar{\mathit{A}}\wedge\bar{\mathit{B}}=\mathrm{span}(\bar{\mathit{A}}\cup\bar{\mathit{B}})
\label{Union}
\end{equation} 
where $\mathit{X}=\mathit{A}\wedge\mathit{B}$.
Geometrically, this means that the wedge of two extensors corresponds to the union of their associated vector spaces. The above equation is the key factor in visualizing these algebraic expressions by linear varieties. The following diagram demonstrates the correspondence between the Grassmann algebra and Grassmann manifold: 
\begin{eqnarray*}
\wedge :\ \mathbb{P}(\bigwedge ^{j}\mathbb{K}^{n})\times\mathbb{P}(\bigwedge ^{k}\mathbb{K}^{n})\ \ \ \longrightarrow\ \ \ \ \ \ \ \ \mathbb{P}(\bigwedge ^{j+k-n}\mathbb{K}^{n})\\
\ \ \ \ \ \ \ (\mathit{A},\mathit{B})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \longmapsto\ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathit{A}\wedge\mathit{B}\ \ \ \ \ \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \uparrow\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \uparrow\ \ \ \ \ \ \ \\
\cup :\ \ \ \ \ \mathit{G}(j,n) \times\mathit{G}(k,n)\ \ \ \ \ \ \longrightarrow\ \ \ \ \ \mathit{G}(j+k-n,n)\\
\ \ \ \ \ \ \ \ \ \ (\bar{\mathit{A}},\bar{\mathit{B}})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \longmapsto\ \ \ \ \ \ \ \ \ \ \ \ \mathrm{span}(\bar{\mathit{A}}\cup\bar{\mathit{B}})\ \ \ \ \ \\
\end{eqnarray*}
Note that the vertical maps are injective and hence assigning linear varieties to $\mathbb{P}(\bigwedge ^{j}\mathbb{K}^{n})$ for different $j$ is uniquely done.
