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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 10 Solutions}
\begin{document}

\maketitle
\begin{enumerate}
    \item Let $V$ be a vector space of dimension $n$ over a field $k$.
    \begin{enumerate}
        \item Define
            $$\phi:V^*\times \bigwedge^n V \longrightarrow \bigwedge^{n-1} V$$
            by
            $$\phi(f,v_1\wedge\cdots\wedge v_n)=\sum_{i=1}^n (-1)^i f(v_i) v_1\wedge\cdots\wedge \hat{v_i}\wedge\cdots\wedge v_n,$$
            where $\hat{v_i}$ denotes the omission of $v_i$ from the wedge product.  We first check that
            $\phi$ is well defined.  For this, it is enough to show that if $x\in \bigwedge^n V$ is 0 then $\phi(f,x)=0$ for any $f\in V^*$.
            We restrict ourselves to the case that $x=v_1\wedge \cdots \wedge v_n$ and $v_1=v_j$ for $1< j\leq n$.  The general
            case is not much more difficult.  We may permute $v_j$ so that it is adjacent to $v_1$, and we only change the sign of
            our wedge product.  Thus, without loss of generality, suppose that $j=2$.  Then
            \begin{align*}
                \phi(f,v_1\wedge v_2\wedge\cdots \wedge v_n)&=f(v_1) v_2\wedge\cdots\wedge v_n-f(v_2) v_1\wedge v_3\wedge\cdots\wedge v_n\\
                &=f(v_1) v_2\wedge\cdots\wedge v_n-f(v_1) v_2\wedge \cdots\wedge v_n\\
                &=0,
            \end{align*}
            where we have used the fact that $v_1\wedge\cdots\wedge \hat{v_i}\wedge\cdots \wedge v_n=0$ for $i\neq 1,2$.

            It is straightforward to verify that $\phi$ is bilinear, so that we have a well defined bilinear map $\phi$.  This
            induces a linear map
            $$\varphi:V^*\otimes \bigwedge^n V \longrightarrow \bigwedge^{n-1} V.$$
            We claim that $\varphi$ is surjective.  Let $e_i$ be a basis for $V$.  Then it suffices to show that the basis
            elements
            $$w_i=e_{1}\wedge\cdots\wedge \hat{e_{i}}\wedge\cdots\wedge e_n$$ of $\bigwedge^{n-1} V$ are in the image of $\varphi$.
            Define $f_{i}\in V^*$ by
            $$f_{i}(e_j)=\begin{cases} 1 & \text{if}\ i=j\\ 0 & \text{otherwise}\end{cases}.$$
            Then we have
            $$\varphi(f_i\otimes e_1\wedge\cdots\wedge e_n)=(-1)^i w_i.$$
            It follows that $\varphi$ is surjective.  Since $\bigwedge^n V$ is one dimensional, and $V^*$ is (non-canonically) isomorphic
            to $V$, and since $\bigwedge^{n-1} V$ is $n$ dimensional, it follows that
            $$\dim(V^*\otimes \bigwedge^n V)=\dim(\bigwedge^{n-1} V),$$
            so that since $\varphi$ is surjective, it must also be injective.  Hence $\varphi$ is an isomorphism, as desired.

        \item Suppose $k<n$ and let $\w\in \bigwedge^{k} V$.  Suppose that $\w\neq 0$.  Pick a basis $e_1,\ldots,e_n$ of $V$
        and let
        $$\w=\sum_{i_1<\ldots<i_k}a_{\overline{i}}e_{i_1}\wedge\cdots\wedge e_{i_k}.$$
        Since $\w$ is nonzero, we must have $a_{\overline{i}}\neq 0$ for some $\overline{i}=(i_1,i_2,\ldots,i_k)$.  Let
        $$\eta=\wedge_{l\not\in \{i_1,i_2,\ldots i_k\}} e_{l}.$$  Then we have
        $$\eta\wedge\w=a_{\overline{i}}\neq 0,$$ so that, in particular,
        $$\w\wedge e_l\neq 0$$ for any $l\not\in\{i_1,\ldots,i_k\}$.
        It follows that if $\w\wedge v=0\in \bigwedge^{k+1} V$ for all $v\in V$ then $\w=0.$

    \end{enumerate}

    \item
    \begin{enumerate}
        \item Define
        $$S^k\r:G\longrightarrow \Gl(S^k V)$$ and
                $$\bigwedge^k\r:G\longrightarrow \Gl(\bigwedge^k V)$$
        by
        $$S^k\r (g)(v_1\otimes \cdots\otimes v_k)=\r(g)v_1\otimes\cdots\otimes\r(g) v_k$$
        and
        $$\bigwedge^k\r (g)(v_1\wedge \cdots\wedge v_k)=\r(g)v_1\wedge\cdots\wedge\r(g) v_k,$$
        respectively, and extend by linearity to all of $S^k V,\ \bigwedge^k V.$
        It is straightforward to verify that these maps are well defined (i.e. that they respect the nontrivial identities
        in $S^k V$ and $\bigwedge^k V$).

        We show that $S^k\r$ is a group homomorphism.  Indeed, we have
        \begin{align*}
            S^k\r(gh)v_1\wedge\cdots\wedge v_n&=\r(gh)v_1\otimes\cdots\otimes\r(gh) v_k\\
            &=\r(g)\r(h)v_1\otimes\cdots\otimes\r(g)\r(h) v_k\\
            &=S^k\r(g)(\r(h)v_1\otimes\cdots\otimes \r(h) v_k)\\
            &=S^k\r(g)\circ S^k\r(h)(v_1\otimes\cdots\otimes v_k),
        \end{align*}
        since $\r$ is a group homomorphism.  The same verification works identically for $\bigwedge^k \r$.
        Observe that
        $$S^k\id=\bigwedge^k \id=\id.$$  Thus we see that the image of $S^k\r$ is indeed in $\Gl(S^k V)$ since
        $$S^k\r(g^{-1})\circ S^k\r (g)=\id.$$
        Similarly for $\bigwedge^k \r$.  It follows that $S^k\r,\ \bigwedge^k\r$ are representations.

        \item Let $\dim V=n$ and realize $\r(g)\in \Gl(V)$ by $n\times n$ matrices.  We have seen that
        $$\bigwedge^n \r(g)=\det (\r(g)),$$
        so that the one dimensional representation on $\bigwedge^n V\simeq k$ given by $\bigwedge\r$
        is multiplication by the determinant of $\r(g)$ (e.g. by the $1\times 1$ matrix $\det(\r(g))$).

        \item Let $e_1,e_2$ be the standard basis of $\C^2$ as a $\C$ vector space and let
        $$s=\begin{pmatrix}a & b \\ c & d \end{pmatrix}\in \Gl_2(\C)$$.  Then we easily have
        \begin{align*}
            s(e_1)&=ae_1+ce_2\\\
            s(e_2)&=be_1+de_2.
        \end{align*}
        It follows that
        \begin{alignat*}{3}
            S^2\id(s)(e_1\otimes e_1)&=(ae_1+ce_2)\otimes (ae_1+ce_2)&=a^2 e_1\otimes e_1+2ac e_1\otimes e_2 +c^2 e_2\otimes e_2\\
            S^2\id(s)(e_1\otimes e_2)&=(ae_1+ce_2)\otimes (be_1+de_2)&=ab e_1\otimes e_1+(ad+bc) e_1\otimes e_2 +cd e_2\otimes e_2\\
            S^2\id(s)(e_2\otimes e_2)&=(be_1+de_2)\otimes (be_1+de_2)&=b^2 e_1\otimes e_1+2bd e_1\otimes e_2 +d^2 e_2\otimes e_2.
        \end{alignat*}
        Thus, in the basis $e_1\otimes e_1,\ e_1\otimes e_2,\ e_2\otimes e_2$
        of $S^2 \C^2$ the matrix of $S^2\id (s)$ is given by
        \begin{align*}
            \begin{pmatrix} a^2 & ab & b^2\\ 2ac & ad+bc & 2bd \\ c^2 & cd & d^2\end{pmatrix}.
        \end{align*}

    \end{enumerate}


    \item
    \begin{enumerate}
        \item Recall that $$\ker g=\{v\in V:g(x,v)=0\ \text{for all}\ x\in V$$
        while $$\ker L_g=\{v\in V:L_g(v)\equiv 0\}=\{v\in V:L_g(v)(x)=g(x,v)=0\ \text{for all}\ x\in V.$$
        Thus it is easy to see that $\ker L_g=\ker g$.  Since $\dim V=n$ it follows from the rank nullity theorem
        that
        $$\rank (L_g)+\dim \ker L_g= \rank(g) +\dim \ker g=n,$$ as required.  Moreover, we know that $g$ is nondegenerate
        if and only if $\ker g=\{0\}$, that is, iff $\dim\ker g=0$, if and only if $\rank(g)=n$.

        \item It suffices to show that there exist symmetric $n\times n$ matrices of any rank $0\leq r\leq n$.  This is clear,
        since
            $$M=\begin{pmatrix}I_r & \\ & 0_{n-r}\end{pmatrix}$$
        clearly has rank $r$ and is symmetric.  Here $I_r$ is the square $r\times r$ identity matrix and $0_{n-r}$ is the square matrix
        of size $n-r$ consisting of all zeroes.

        Now let $g$ be an alternating form on $V$ and set $N=\ker g$.  Then $h=g|W$ is nondegenerate and skew, where
        $$W=V/N$$ and
        $$\rank(h)=\rank(g)=\dim W.$$
        Now $h^T=-h$ so that
        $$\det(h)=\det(h^T)=\det(-h)=(-1)^{\dim W}\det(h)$$
        from which it follows that $\dim W=\rank g$ is even since $\det h\neq 0$.

        As above, it is enough to show that there exist skew symmetric $n\times$ matrices of any given even rank.  Let
                    $$M=\begin{pmatrix}T &  &  & \\ & T & & \\ & & \ddots & \\ & & & 0_{n-2r}\end{pmatrix}$$
        where $$T=\begin{pmatrix}0 & -1\\ 1 & \phantom{-}0\end{pmatrix}$$ and there are $r$ $T$'s along the diagonal.
        This matrix clearly has rank $2r$ and is skew.

        \item Let $W\subset V$ be a codimension 1 subspace and suppose that $g$ is a nondegenerate symmetric form on $V$.
        Then
        $$L_g:V\rightarrow V^*$$ is an invertible matrix.  Moreover, since $\dim W=n-1$ we see that
        $$\dim(L_g(W))=n-1$$ where $L_g(W)\subset V^*$.  The rank of $g|W$ is the rank of $L_{g|W}$, which
        is the dimension of
        $$L_g(W)\cap W^*.$$  Since $W^*$ also has dimension $n-1$ and $V$ has dimension $n$, the intersection
        $L_g(W)\cap W^*$ has dimension at least $n-2$.  Thus, the possibilities for the rank of $g|W$ are
        $$n-1,\ n-2.$$  It is not difficult to find examples where each possibility occurs.  Indeed,
        The $\rank(g_W)=n-1$ if and only if $L_g(W)=W^*$.

        \item Set
        $$g(v,w)=\sum_{i=1}^{n} v_i w_{n-i+1}$$ and let $W=\{v\in V: v_i=0\ \text{for}\ 1\leq i\leq m\}$.  Then it is easy to
        see that $g$ is symmetric and $g(w_1,w_2)=0$ for any $w_1,w_2\in W$ but that for any $w\neq 0\in W$ there exists $v\in V\setminus W$ such that
        $g(w,v)\neq 0$.  It follows that $g$ has rank $m$.
        Similarly, the nondegenerate skew symmetric form
        $$g(v,w)=\sim_{i=1}^m (v_i w_{n-i+1}-v_{n-i+1}w_i)$$ has the $m$ dimensional isotropic subspace
        $$W=\{v\in V: v_i=0\ \text{for}\ 1\leq i\leq m\}$$.

        Finally, it is not difficult to see that the usual dot product on $\R^n$ has no isotropic subspaces other than $\{0\}$.
        This follows from the fact that the only solution to
        $$\sum_{i=1}^n v_i^2$$ over $\R$ is $v=0$.

    \end{enumerate}

\end{enumerate}

\end{document}
