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\define\GL{\operatorname{GL}}
\define\ORTH{\operatorname{O}}
\define\Hom{\operatorname{Hom}}
\define\End{\operatorname{End}}
\define\Aut{\operatorname{Aut}}
\define\Ker{\operatorname{Ker}}
\define\Coker{\operatorname{Coker}}
\redefine\Im{\operatorname{Im}}
\define\Rank{\operatorname{Rank}}
\define\R{\Bbb R}
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\title The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 12
\endtitle
\author Professor Steven Bradlow \\ Class Notes From Math 433
\endauthor
\affil University of Illinois at Urbana-Champaign
\endaffil
\address 273 Altgeld Hall, 1409 W. Green Street, Urbana, IL 61801
\endaddress
\email bradlow\@math.uiuc.edu
\endemail
\date February 18, 1998
\enddate
\endtopmatter
\document
We will need to be familiar with Grassmanians and bundles over them
for our construction of the unviversal bundle.
\vskip.25in
\definition{Definition 1} The real {\it Grassmanian} is the set of all
$k-$planes in $\R^n$ and is denoted by $G_\R(k,n)$. Similarly, the
complex {\it Grassmanian}, $G_\C(k,n)$ is the set of all complex $k-$planes
in $\C^n$.
\enddefinition
\vskip.25in
\example{Example 1} $G(1,n)$ is the set of all lines in $\Bbb R^n$. This
should be familiar as real projective $n-1$ space, $\rp^{n-1}$. Similarly,
$G_\C(1,n)=\cp^{n-1}$.
\endexample
\vskip.25in
We need to establish the following key properties of the Grassmanian. We
derive the properties for $G(k,n)=G_\R(k,n)$ but the same hold for $G_\C(k,n)$. These
are:
\vskip.25in
\itemitem{(a)} $G(k,n)$ is a smooth manifold of dimension $k(n-k)$.
\vskip.25in
\itemitem{(b)} Let $\gamma_k^n$ be the subset of $G(k,n) \times \Bbb R^n$ consisting of $\{(V, \vec v):\vec v \in V\}$. Then under the natural projection map, $\pi:\gamma_k^n \to G(k,n)$ is a bundle with fiber a $k-$plane.
\vskip.25in
Recall that the trivial bundle is denoted by $\underline{\Bbb R^n}$. Then the universal quotient bundle, $Q$, over $G(k,n)$ is defined
by the exact sequence $$0 \to \gamma_k^n \to \underline{\Bbb R^n} \to Q \to 0$$
We have that $Q_{|V} \cong \Bbb R^n/V$, an $n-k$-plane.
\vskip.25in
We begin with the first property above. A useful was to specify a frame is by
tge $n \times k$ matrix, $[\vec v_1,\dots,\vec v_k]$ (where eacg $\vec v_i$ is
displayed as an $n \times 1$ column). The linear independence condition is
equivalent to the condition that $[\vec v_1,\dots,\vec v_k]$ has rank $k$.
We can interpret this matrix as an injective linear map, $\R^k \to \R^n$.
Of course each $k-$plane admits may frames, so that the description is in
no way unique. In fact, given any $A \in \GL(k)$, we get a new frame described
by the matrix, $[\vec v_1,\dots,\vec v_k] \cdot A$, or equivalently, the
map
$$\harrowlength=.42\sarrowlength
\commdiag{\R^k \mapright^A &\R^k \mapright^{[\vec v_1,\dots,\vec v_k]} &\R^n}$$
This leads to a description of $G(k,n)$ as a quotient (by $\GL(k)$) of the
space of all {\it k-frames}. The latter is called the Stiefel manifold of
$k-$frames and is denoted by $F(k,n)$. We can identify
$$\aligned F(k,n)&=\{n \times k \text{ matrices of rank} k\} \cr
&=\{f:\R^k \to \R^n:f \text{ is linear and injective}\} \endaligned $$
Then, $G(k,n)=F(k,n)/\GL(n)$ where $\GL(k)$ acts on $F(k,n)$ in the manner
described above.
\vskip.25in
\remark{Exercise 1} Show that the $\GL(k)$ action on $F(k,n)$ is a free right action.
\endremark
\remark{Exercise 2} Show that $F(k,n)$ is a smooth manifold of dimension $kn$.
\endremark
\vskip.25in
Suppose that $[\vec v_1,\dots,\vec v_k] \in G(k,n)$ has the property that the first $k$ by $k$ minor has non-zero determinant (since the matrix has rank
$k$, at least one of the minors has non-zero determinant). Then, there is
some $A \in \text{GL}(k)$ such that $$[\vec v_1,\dots,\vec v_k] \cdot A=[\frac{\text{Id}}{B}]$$ That is, the first $k$ rows have been reduced to
the identity, and $B$ is an $(n-k)$ by $k$ block. We define a map,
$$\text{Mat}(n-k,k) \to U_{1,\dots,k} \subset \text{GL}(k,n)$$ by
$$B \mapsto [\frac{\text{Id}}{B}]$$
\vskip.25in
\remark{Exercise 3} If $I=\{i_1,\dots,i_k\} \subseteq \{1,\dots,n\}$, then $U_I$ defines a coordinate patch for $G(k,n)$. Show that $G(k,n)$ is covered
by the set of all such $U_I$ and that the coordinate transformations are
smooth.
\endremark
\vskip.25in
Note that the story gets better. Over $\C^n$, the coordinate transformations for $G_\C(k,n)$ are holomorphic. Thus, $G_\C(k,n)$ is a complex manifold
of dimension $k(n-k)$.
\vskip.25in
We now move to the second property. We first look at the projection $\pi:F(k,n) \to G(k,n)$. This clearly has fiber $\GL(k)$. In fact, it is a
$\GL(k)$ bundle. We need to establish the local triviality over $U_I$. Over
$U_{\{1,\dots,k\}}$, describe $$[\vec v_1,\dots,\vec v_k]=[\frac{\text{Id}}{B}]\cdot A$$ Define a map $[\vec v_1,\dots \vec v_k] \mapsto ([\vec v_1,\dots,\vec v_k],A)$. This is the trivializing map.
So, $\pi:F(k,n) \to G(k,n)$ is a principal $\text{GL}(k)$-bundle. How does
this bundle relate to $\gamma_k^n$?
\vskip.25in
\proclaim{Claim} $$\gamma_k^n=F(k,n) \times_{GL(k)} \Bbb R^k$$
\endproclaim
\vskip.25in
\remark{Exercise 4} Establish this for the case where $k=1$. the tautological bundle over $\rp^{n-1}$.
\endremark
\vskip.25in
\remark{Remark} Given a $k-$plane, $V^k \subseteq \Bbb R^n$, we obtain an $n-k$ plane given by $\Bbb R^n/V^k$ (if we have a metric, then we can identify this
with $V^\perp$m the othogonal complement of $V$).
\endremark
\vskip.25in
We thus get a map $q:G(k,n) \to G(n-k,n)$. We can use this map to pull back $Q \to G(n-k,n)$ to $G(k,n)$.
We then have the following diagram:
$$\harrowlength=.42\sarrowlength
\commdiag{\gamma_k^n &&q^*(Q) &&Q \cr
&\arrow(1,-1)&\mapdown &&\mapdown \cr
&&G(k,n) &\mapright_{q} &G(n-k,n) }$$
A natural question to ask is: how are $\gamma_k^n$ and $q^*(Q)$ realted. It turns out that $\gamma_k^n$ is the same as the dual bundle to $q^*(Q)$.
\vskip.25in
\remark{Exercise 5} Prove that $\gamma_k^n \cong (q^*(Q))^*$.
\endremark
\enddocument
\end