text:
\section{Introduction}
Recently there has been an increasing interest both on ideals failing Lefschetz Pro\-perties and on arrangements of hyperplanes. In \cite{DIV} the authors link these two topics. In fact, given a suitable line arrangement in the projective plane, i.e. a finite collection of lines, they relate the unstability of its derivation bundle (see Section \ref{arrang_sec}) to the failure of the Strong Lefschetz Property at range 2  of the ideal generated by suitable powers of the linear forms defining the lines (see Section \ref{lef_sec}). This linkage is deepened in \cite{CHMN16} where the authors add the equivalent condition of existence of a suitable unexpected curve.
\\ Here we recover these arguments in order to improve and generalize \cite{DIV}. Precisely, in Section \ref{lef_sec} we generalize the main theorem in \cite{DIV} to the case when there exist syzygies of suitable degree (Theorem \ref{thgen}), and in Section \ref{arrang_sec} we reformulate and improve some results in \cite{DIV}. More precisely, given an artinian ideal $I$ ge\-nerated by powers of linear forms, we characterize geometrically the existence of syzygies of degree $0$. Let $Z$ be
the set of points which are dual to these linear forms, then the existence of $0$-syzygies in $I$ is related to the number of aligned points in $Z$ (Theorem \ref{th_aligned}). In this way we restate some results (Proposition \ref{prop_bundle} and Conjecture \ref{conj_terao}) in a more geometric form. The note ends with the equivalence among the existence of unexpected curves, the failure of SLP at range 2 and the existence of Laplace equations in suitable cases (Corollary \ref{cor_unexp}).

\section{Lefschetz properties}\label{lef_sec}
Let $\mathbb K$ be an algebraically closed field of characteristic $0$, $R=\mathbb K[x_0,\ldots, x_n]=\bigoplus R_t$ be the graded polynomial ring in $n+1$ variables over $\mathbb K$ and $r_t= \dim R_t=\dim H^0(\mathscr O_{\check{\mathbb P}} \newcommand{\G}{\mathbf G^n}(t))=\left( \begin{array}{c}t+n\\n\end{array}\right)$, when $n$ is fixed.
\\Let
$$A=R/I= \bigoplus_{i=0}^{m}A_i$$ be a graded artinian algebra, defined by a homogeneous ideal $I$. Note that $A$ is finite dimensional over $\mathbb K$. We denote by $H_A=H_{R/I}$ the Hilbert function of $R/I$.
\\First of all, we recall the definition of algebra (or ideal) failing the Lefschetz Properties.
\begin{defi}
 The artinian algebra $A$ (or the artinian ideal $I$) fails the Weak Lefschetz Property (from now on WLP) if  for any linear form $L$   there exists $i$  such that the multiplication map by $L$,
$ \times L : A_i \rightarrow A_{i+1},$
 does not have maximal rank (i.e. is neither injective nor surjective). More precisely, $A$ (or $I$) fails WLP \em by $\delta$\em\ if the multiplication map has rank $\min\{H_{R/I}(i),H_{R/I}(i+1)\}-\delta$(see \cite{CN13}).\end{defi}
Similarly, we define the failure of the Strong Lefschetz Property by $\delta$.
\begin{defi} The artinian algebra $A$ (or the artinian ideal $I$)  fails the Strong
Lefschetz Property (from now on SLP) at range $k$ and degree $i$ by $\delta$ (with $\delta \geq 1$) if  for any linear form $L$  the multiplication map $ \times L^k : A_i \rightarrow A_{i+k},$ has rank $\min\{H_{R/I}(i),H_{R/I}(i+k)\}-\delta$.
\end{defi}
One of the main examples comes from a classical result of Togliatti: the ideal $I=(x^3,y^3,z^3,xyz)$ fails the WLP in degree $2$ by $1$. In \cite[Example 3.1]{BK} this ideal is studied, but it appeared in \cite{T} in terms of projection center of the Veronese surface (see also \cite{DI} for a modern approach and below for some details).
\\
In \cite{DIV} the authors of this note and J. Vall\`{e}s characterize artinian ideals failing SLP at range $k$ by $\delta$ in terms of suitable projections of Veronese varieties satisfying Laplace equations and of the existence of suitable singular hypersurfaces. In order to state our main result we recall some definitions and results from \cite{DIV}.
\begin{defi}
Let $I=(F_1, \ldots, F_r) \subset R$ be an artinian ideal generated by
forms of degree $d$. The syzygy bundle $K$ is defined by the exact sequence
$$  \begin{CD}
 0@>>> K @>>>  \mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^{n}}^{r} @>\Phi_{I}>>
 \mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^{n}}(d) @>>> 0,
\end{CD}$$
where $\Phi_{I}(a_1,\ldots,a_r)=a_1F_1+\ldots+a_rF_r.$
\end{defi}
\begin{thm}\cite[Theorem 4.1]{DIV}
\label{p1}
 Let $I=(F_1, \ldots, F_r) \subset R$ be an artinian ideal generated by
forms of degree $d$ and  $K$ the syzygy bundle.
Let $i$ be a non-negative integer such that $ \mathrm{h}^0( K(i))=0$ and $k$ be an integer such that $k\ge 1$.
Then $I$ fails the SLP at the range  $k$ in degree $d+ i-k$  if and only if the induced homomorphism on global sections
(denoted by $\mathrm{H}^0(\Phi_{I,L^k})$)
$$  \begin{CD}
  \mathrm{H}^0(  \mathscr O _{L^k}(i))^r @>\mathrm{H}^0(\Phi_{I,L^k})>>
 \mathrm{H}^0(  \mathscr O _{L^k}(i+d))
\end{CD}$$
does not have maximal rank for a general linear form $L$.
\end{thm}

\begin{defi}
 Let $X\subset \mathbb P^N$ be a projective $n$-dimensional complex variety. For $m\geq 1$, the projective $m$-th osculating space to $X$ at a general point $P$, $T_P^{m}(X)$, is the subspace of $ \mathbb P^N$ spanned by $P$ and by all the derivative points of order less  than or equal to $m$ of a local parametrization of $X$, evaluated at $P$. Of course, for $m=1$ we get the tangent space $T_P(X)$.
\end{defi}
\noindent We remark that the expected dimension of the $m$-th osculating space is $$ \mathrm{expdim} T_P^m(X)=\mathrm{inf}\left(\binom{n+m}{n}-1,N\right).$$
\begin{defi}A $n$-dimensional variety $X\subset \mathbb P^N$  satisfies  $ \delta$ independent Laplace equations of order $m$ if the $m$-th osculating space at a general point has $$\dim T_P^{m}(X)=
\mathrm{expdim} T_P^m(X)-\delta.$$\\
If $N < \binom{n+m}{n}-1$, then there are always $\binom{n+m}{n}-1-N$ relations between the partial derivatives. We call these relations \lq\lq trivial\rq\rq\ Laplace equations of order $m$. \end{defi
\par\vspace{3mm} Among the varieties satisfying non trivial Laplace equations, we are inte\-rested in rational varieties which are suitable projections of a Veronese variety.
\\ For any vector space $V$, $V^*=\mathrm{Hom}_{\C}(V,\C)$ will be the dual space.\\
 Let $v_{t} : [L] \in \mathbb P (R_1^*)\hookrightarrow [L^{t}]\in \mathbb P (R_{t}^*)$
be the $t$-uple Veronese embedding whose image  $v_t(\mathbb P^n)$ is the Veronese $n$-fold of order $t$.
\\Let $I=(F_1, \ldots, F_r)$  be an ideal generated by $r$ forms of degree $d$ and $I_h$ be the homogeneous component of degree $h$ of $I$ for any $h$.
\begin{defi}The apolar space of $I$ in degree $d+i$ with $i\geq 0$ is
$$  I_{d+i}^{\perp}=\{\Delta \in R_{d+i}^*\,|\,\Delta(F)=0, \,\, \forall F\in I_{d+i}\},$$ where the canonical basis of $R_{d+i}^*$ is given by the $r_{d+i}=\binom{d+i+n}n$ derivations $ \frac{\partial^{d+i}}{\partial x_0^{i_{0}}\ldots \partial x_n^{i_n}}$ with $ i_0+\ldots +i_n=d+i.$ \end{defi
There is the exact sequence of vector spaces
$$  \begin{CD}
0 @>>> I_{d+i}^{\perp} @>>>R_{d+i}^* @>>> I_{d+i}^* @>>> 0
\end{CD}$$
and, by dualizing it, one can identify $R_{d+i}/I_{d+i}\simeq (I_{d+i}^{\perp})^*$ and write the decomposition
$R_{d+i}=I_{d+i} \oplus (I_{d+i}^{\perp})^*.$
\\ By denoting the corresponding  projection map
$$\begin{CD}
\pi_{I_{d+i}}: \mathbb P(R^*_{d+i})\setminus \mathbb P(I^*_{d+i}) \rightarrow \mathbb P(I^\perp _{d+i})

  \end{CD},
$$ we consider the variety $X:=\pi_{I_{d+i}}(v_{d+i}(\mathbb P^n))$.
\begin{rem}
The toric case is the easiest one: when $I_d$ is generated by $r$ monomials of degree $d$,   $(I_{d}^{\perp})^*$ is  generated by
the  remaining $r_d-r$ monomials.
\\ The case of powers of linear forms is very interesting. In \cite{EI} it is proved that the apolar of an ideal generated by powers of linear forms is related to the $0$-dimensional scheme of the dual points of the linear forms, as in the following Theorem.
\begin{thm}\cite{EI}\label{EI}
Let $l_1, \ldots, l_r$ be linear forms in $\mathbb P^n$, $d_1,\ldots,d_r >0$ be integers and $I=(l_1^{d_1},\ldots,l_r^{d_r})$. If $P_i=l_i^\vee$ denotes the dual point of $l_i$, then for any $j\geq \max{d_i}$,
$$\dim_{\mathbb K}\left(\frac R I\right)_j = \dim_{\mathbb K} \bigcap_{i=0,\ldots,r}I_{P_i}^{j-d_i+1}.$$
\end{thm}
\end{rem}
\begin{nota} For any $i\ge 0$, $k\ge 1$, we denote $N(r,i,k,d):=r(r_i-r_{i-k})- (r_{d+i}-r_{d+i-k})$,
$N^{+}=\mathrm{sup}(0,N(r,i,k,d))$ and $N^{-}=\mathrm{sup}(0,-N(r,i,k,d)).$
\end{nota}
\begin{rem}\label{remark}
Note that when $\mathrm h^0 (K(i))=0$ then $N(r,i,k,d)=H_{R/I}(d+i-k)-H_{R/I}(d+i)=\dim \ker (\times L^k) - \dim \mathrm{coker} (\times L^k)$.
\\Moreover if $N < \binom{n+d+i-k}{n}-1$ then the number of trivial equations is exactly
$N^+ + \dim I_{d+i-k}$.
\end{rem}
The main result in \cite{DIV} (that generalizes to SLP the result given in \cite{MMO} on WLP) is the following.
\begin{thm}\cite[Theorem 5.1]{DIV}
\label{th1bis}
Let $I=(F_1, \ldots, F_r) \subset R$ be an  artinian ideal generated by $r$ homogeneous polynomials  of degree $d$.
Let $i,k,\delta$ be integers such that
i\ge 0$, $k\ge 1$.
Assume that there is no syzygy  of degree $i$  among the $F_j$'s.
The following conditions are equivalent:
\begin{enumerate}
\item The ideal $I$ fails the SLP at the range  $k$ in degree $d+ i-k$.
\item For a general linear form $L$ of $\check{\mathbb P}} \newcommand{\G}{\mathbf G^n$, there exist  $N^{+}+\delta$, with $\delta \ge 1$,  independent vectors  $(G_{1j},\ldots, G_{rj})_{j=1, \ldots,N^{+}+\delta }\in R_i^{\oplus r}$ and
$N^{+}+\delta $ forms $G_j\in R_{d+ i-k}$ such that
$G_{1j}F_1+ \ldots + G_{rj}F_r=L^kG_j$.
\item
 The $n$-dimensional variety $\pi_{I_{d+i}}(v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n))$ satisfies
$\delta \ge 1$ Laplace equations of order $d+i-k$.
\item \label{item_iv_thm} For any  $L\in R_1$, $\mathrm{dim}_{\C}((I_{d+i}^{\perp})^*\cap \mathrm{H}^0(\mathcal{I}_{L^{\vee}}^{d+i-k+1}(d+i))\ge N^{-}+\delta$, with $\delta \ge  1$.
\end{enumerate}
\end{thm}
 As noted in \cite{DIV} the hypothesis on the global syzygy in Theorem \ref{th1bis} is not restrictive in case of syzygies of degree $i\geq 1$.
\begin{lem}\cite[Lemma 5.2]{DIV}
 \label{lem-syz}
Let $I$ be the ideal $(L_1^{d},\ldots,L_r^{d})$  where the $L_j$ are general linear forms and $r<r_d$. Let $K$ be its syzygy bundle and $i\geq 1$. Then
$$ \hh^0(K(i))=0 \Leftrightarrow rr_i\le r_{d+i}.$$
\end{lem}
 Here we generalize these results in the case in which $H^0(K(i))\neq 0$ and $H^0(K(i-k))= 0$. This generalization is interesting as if there are $0$-syzygies, i.e. $i=0$ and $H^0(K)\neq 0$ then the condition $H^0(K(-k))= 0$ is automatically verified.
\begin{thm}
\label{p1bis}
Let $i,k$ be non-negative integers such that $\mathrm{h}^0( K(i))=s$ and $H^0(K(i-k))= 0$.
Denoted by $\mathrm{H}^0(\Phi_{I,L^k})$ the induced homomorphism on global sections
$$  \begin{CD}
  \mathrm{H}^0(  \mathscr O _{L^k}(i))^r @>\mathrm{H}^0(\Phi_{I,L^k})>>
 \mathrm{H}^0(  \mathscr O _{L^k}(i+d))
\end{CD},$$
then the homomorphisms $\mathrm{H}^0(\Phi_{I,L^k})$ and  $\times L^k$ have the same cokernel, while $\ker (\times L^k)\cong \dfrac{\ker \mathrm{H}^0(\Phi_{I,L^k})}{\mathrm{H}^0( K(i))}$.
\end{thm}
\begin{proof} The proof exploits the same ideas as in \cite[Theorem 4.1]{DIV}.
Let us consider the canonical exact sequence
$$  \begin{CD}
 0@>>> K(i-k) @> \times L^k>> K(i) @>>>
 K\otimes \mathscr{O}_{L^k}(i) @>>> 0.
\end{CD}$$
As $H^0(K(i-k))= 0$ and $A_{d+i}=\mathrm{H}^1(K(i))$ for any $i\in \mathbb Z} \newcommand{\C}{\mathbb K$ (\cite[Proposition 2.1]{BK}), we obtain the long exact sequence of cohomology
$$
0 \rightarrow \mathrm{H}^0( K(i))\rightarrow
  \mathrm{H}^0(K\otimes \mathscr O _{L^k}(i)) \rightarrow A_{d+ i-k} \stackrel{\times L^k}\longrightarrow  A_{d+i} \rightarrow$$$$\rightarrow
\mathrm{H}^1(K\otimes \mathscr O _{L^k}(i))\rightarrow \mathrm{H}^2(K(i-k)) \rightarrow 0.
$$
Moreover, by tensoring the exact sequence defining the bundle $K$  by $\mathscr O _{L^k}(i)$
$$  \begin{CD}
 0@>>> K @>>>  \mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^{n}}^{r} @>\Phi_{I}>>
 \mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^{n}}(d) @>>> 0,
\end{CD}$$
we get in cohomology
$$ \begin{array}{ll} 0\longrightarrow &
  \mathrm{H}^0(K\otimes   \mathscr O _{L^k}(i)) \longrightarrow \mathrm{H}^0(  \mathscr O _{L^k}(i))^r \stackrel{\mathrm{H}^0(\Phi_{I,L^k})}\longrightarrow
 \mathrm{H}^0(  \mathscr O _{L^k}(i+d))\longrightarrow  \\ & \mathrm{H}^1(K\otimes   \mathscr O _{L^k}(i))\longrightarrow \mathrm{H}^1(  \mathscr O _{L^k}(i))^r \longrightarrow
 \mathrm{H}^1(  \mathscr O _{L^k}(i+d))
\longrightarrow 0.\end{array}$$
\\ Moreover $\mathrm{H}^2(K(i-k))=0=\mathrm{H}^1(\mathscr O _{L^k}(i))=0$ when $n>2$; while when $n=2$ we have $\mathrm{H}^2(K(i-k))=\mathbb K^t$ with $t= rr_{k-i-3} - r_{k-i-d-3}$ and
 $\mathrm{h}^1(\mathscr O _{L^k}(i)) =\mathrm{h}^2(\mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^2}(i-k))=r_{k-i-3}$,  $\mathrm{h}^1(\mathscr O _{L^k}(i+d)) =\mathrm{h}^2(\mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^2}(i+d-k))=r_{k-i-d-3}$;
 so the cokernel of both maps $\mathrm{H}^0(\Phi_{I,L^k}) $ and $\times L^k$ are the same and between the kernels there is the sought isomorphism.
\end{proof}
Following Remark \ref{remark}, it is natural to generalize the integers $N(r,i,k,d), N^+,N^-$ in case of syzygies.
\begin{defi}\label{def_Ns}
Let $h^0(K(i))=s$, we define:\begin{itemize} \item $N_s=N(r,i,k,d,s):=r(r_i-r_{i-k})- (r_{d+i}-r_{d+i-k})-s$; \item $N^{+}_s:=\mathrm{max}(0,N_s)$; \item $N^{-}_s:=\mathrm{max}(0,-N_s)).$\end{itemize}
\end{defi}
\begin{rem} Note that when $h^0(K(i-k))=0$ we have $N_s=H_{R/I}(d+i-k)-H_{R/I}(d+i)$.
\end{rem}
Now we give the main theorem of this note, that generalizes \cite[Theorem 5.1]{DIV}.
\begin{thm}
\label{thgen}
Let $I=(F_1, \ldots, F_r) \subset R$ be an  artinian ideal generated by $r$ homogeneous polynomials  of degree $d$.
Let $i,k,\delta$ be non negative integers such that there is no syzygy  of degree $i-k$  among the $F_j$'s.
The following conditions are equivalent:
\begin{enumerate}
\item \label{item_SLP}The ideal $I$ fails the SLP at the range  $k$ in degree $d+ i-k$ by $\delta$;
\item\label{item_ker} $\dim \ker (\times L^k)=N^+_s+\delta>N^+_s$;
\item\label{item_coker} $\dim \mathrm{coker} (\times L^k)=N^-_s+\delta>N^-_s$;
\item \label{item_equation}The $n$-dimensional variety $\pi_{I_{d+i}}(v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n))$ satisfies
$\delta \ge 1$ non trivial Laplace equations of order $d+i-k$ and no Laplace equation of smaller order;
\item \label{item_hypersurf} For any  $L\in R_1$, $\mathrm{dim}_{\C}((I_{d+i}^{\perp})^*\cap \mathrm{H}^0(\mathcal{I}_{L^{\vee}}^{d+i-k+1}(d+i))= N^{-}_s+\delta$, with $\delta \ge  1$.
\end{enumerate}
\end{thm}
\begin{proof}
By Theorem \ref{p1}, the equivalence between \eqref{item_ker} and \eqref{item_coker} is an easy calculation.
Moreover, it is obvious that \eqref{item_ker} and \eqref{item_coker} imply \eqref{item_SLP}.
\\ Let us assume that $I$ fails the SLP at the range  $k$ in degree $d+ i-k$. Let $D=\dim \ker (\times L^k)$ and suppose that $0<D\leq N^+_s$. Denoting $h^0=\dim \ker \mathrm{H}^0(\Phi_{I,L^k})$, by Theorem \ref{p1}, $D=h^0-s$ and $\dim \mathrm{coker} (\times L^k)=D-N_s$. Now, if $N_s\geq 0$, then $N^+_s=N_s=N-s$ and $0<D=h^0-s\leq N-s$ so $h^0\leq N$. On the other hand, $0<\dim \mathrm{coker} \times L^k=D-N_s=h^0-s-N+s=h^0-N$ so $h^0>N$ that is a contradiction.
If $N_s<0$, then $N^+_s=0$ and $0<D\leq 0$ is a contradiction. So \eqref{item_SLP} $\Leftrightarrow$ \eqref{item_ker} $\Leftrightarrow$ \eqref{item_coker}.
\\\eqref{item_ker} $\Leftrightarrow$ \eqref{item_equation}.
 The dimension of the kernel of the map $\times L^k$ i.e. the dimension of $\mathrm{H}^0(K\otimes \mathscr O _{L^k}(i))-s$, written in a geometric way, is
$$N^+_s+\delta=\mathrm{dim}[\check{\mathbb P}} \newcommand{\G}{\mathbf G(I_{d+i}^*) \cap T_{[L^{d+i}]}^{d+i-k}v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n)]+1,\,\,\,\,\, \delta \ge 0$$
where the projective dimension is $-1$ if the intersection is empty.  The number $\delta$ is the number of (non trivial) Laplace equations. Indeed,
 the dimension of the $(d+i-k)$-th osculating space to $\pi_{I_{d+i}}(v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n))$ is $r_{d+i-k} -N^{+}_s -\delta$, since
 the $(d+i-k)$-th osculating space to $v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n)$  meets the center of projection along a $\check{\mathbb P}} \newcommand{\G}{\mathbf G^{N^{+}_s+\delta-1}$.  In other words, the $n$-dimensional variety  $\pi_{I_{d+i}}(v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n))$
 satisfies $\delta$ Laplace equations.
\\ \eqref{item_equation}$\Leftrightarrow$ \eqref{item_hypersurf}.
The image  by $\pi_{I_{d+i}}$ of the  $(d+i-k)$-th osculating space to the Veronese
$v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n)$ at a general point  has codimension $\mathrm{h}^0(K\otimes \mathscr O _{L^k}(i))-N^+_s$ in
$\check{\mathbb P}} \newcommand{\G}{\mathbf G(I_{d+i}^{\perp})$. The codimension corresponds to the number of hyperplanes in $\check{\mathbb P}} \newcommand{\G}{\mathbf G(I_{d+i}^{\perp})$ containing the osculating space to
$\pi_{I_{d+i}}(v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n))$. These hyperplanes are images by $\pi_{I_{d+i}}$ of hyperplanes in $\check{\mathbb P}} \newcommand{\G}{\mathbf G(R_{d+i}^*)$ containing
$\check{\mathbb P}} \newcommand{\G}{\mathbf G(I_{d+i}^*)$ and the $(d+i-k)$-th osculating space to
$v_{d+i}(\check{\mathbb P}} \newcommand{\G}{\mathbf G^n)$ at the point $[L^{d+i}]$.
In the dual setting it means that these hyperplanes define
forms  of degree $d+i$ in $(I_{d+i}^{\perp})^*$ with multiplicity  $(d+i-k+1)$ at $[L^{\vee}]$.
\\
To summarize, the number of non trivial Laplace equations is $\hh^0(K\otimes \mathscr O _{L^k})-s-N^+_s$ and
$\mathrm{coker}(\mathrm{H}^0(\Phi_{I,L^k}))\simeq (I_{d+i}^{\perp})^*\cap \mathrm{H}^0(\mathcal{I}_{L^{\vee}}^{d+i-k+1}(d+i)).$
\end{proof}

\section{Remarks on failing SLP at the range  $2$ and line arrangements on $\check{\mathbb P}} \newcommand{\G}{\mathbf G^2$}
\label{arrang_sec}
In this section we explain the connection with certain line arrangements in the plane and we add some remarks on \cite[Section 7]{DIV}.
\par
A line arrangement  is a collection of distinct lines in the projective plane. Arrangements of lines, and more generally of hyperplanes, have long been an important topic of study (see \cite{Cartier} or \cite{OT} for a good introduction).
\begin{defi}
Given a line arrangement, let us denote by
$f=0$ the equation of the union of lines of this arrangement. The
vector bundle $\mathcal{D}_0$ defined as the kernel of the  jacobian map:
$$ \begin{CD}
    0 @>>> \mathcal{D}_0 @>>> \mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^2}^{3} @>(\partial f)>> \mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^2}(d-1)
   \end{CD}
$$
 is called \em derivation bundle\em\ (or  logarithmic bundle ) of the line arrangement  (see \cite{S}  and \cite{Sc} for an introduction to
derivation bundles).
\\
One can consider the  lines of the arrangement in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^{2}$ as a set of distinct points $Z$  in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^{2\vee}$. Then we will
denote by  $\mathcal{D}_0(Z)$ the associated derivation bundle.

The arrangement of lines is said {\it free with exponents} $(a,b)$ if its derivation bundle splits on $\check{\mathbb P}} \newcommand{\G}{\mathbf G^2$ as a sum of two line bundles, more precisely if
$$ \mathcal{D}_0(Z)=\mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^2}(-a)\oplus \mathscr O _{\check{\mathbb P}} \newcommand{\G}{\mathbf G^2}(-b),$$
while $(a,b)$ is called the \em general splitting type\em\ if it corresponds to the splitting of $\mathcal{D}_0(Z)$ over a general line $l\subset \check{\mathbb P}} \newcommand{\G}{\mathbf G^2$.
\end{defi}
The general splitting type $(a,b)$ is related to the existence of curves of degree $a+1$ passing through $Z$, having multiplicity $a$ at $l^{\vee}\in \check{\mathbb P}} \newcommand{\G}{\mathbf G^{2\vee}$.
More precisely,
\begin{lem}(\cite{V_AnnTol}, \cite[Proposition 2.1]{FV})
\label{linksd} Let $Z\subset \check{\mathbb P}} \newcommand{\G}{\mathbf G^{2\vee}$ be a set of $a+b+1$ distinct points with $1\le a\le b$  and $l$ be
a general line in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^{2}$. Then  the following conditions are equivalent:
\begin{enumerate}
 \item $\mathcal{D}_0(Z)\otimes
\mathscr O _{l}=\mathscr O _{l}(-a)\oplus \mathscr O _{l}(-b)$.
 \item  $\mathrm{h}^0((\mathcal{J}_{Z}\otimes \mathcal{J}_{l^{\vee}}^{a})(a+1))\neq
0$ and $\mathrm{h}^0((\mathcal{J}_{Z}\otimes \mathcal{J}_{l^{\vee}}^{a-1})(a))=
0.$
\end{enumerate}
\end{lem}
So the general splitting type is related to the degree of suitable singular curves through $Z$ and, thanks to the result of Emsalem-Iarrobino (\cite{EI}) here stated in Theorem \ref{EI}, in
\cite[Proposition 7.2]{DIV} an equivalence between unstability of the derivation bundle and the failing of SLP at range $2$ is given. Actually, in the statement of Proposition 7.2 and its corollaries in \cite{DIV} the hypothesis on the non-existence of $0$-syzygies, that is implicitly used in the proofs, is missing. Here we give a precise statement, in a more general form with respect to the number of lines and the splitting type of the derivation bundle. In this way we determine an interval of possible degrees of generators of ideals failing SLP.
\begin{prop}\label{prop_bundle}
\label{th5}
 Let $I\subset R=\C[x,y,z]$ be an  artinian ideal generated by $2d+1+n$  polynomials $l_1^d, \ldots, l_{2d+1+n}^d$ where $n\geq 0$ and $l_i$ are distinct linear forms in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^2$.
Let  $Z=\{l_1^{\vee},\ldots, l_{2d+1+n}^{\vee}\}$ be the corresponding set of points in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^{2\vee}$. If the ideal is minimally generated in degree $d$, then the following conditions are equivalent:
\begin{enumerate}
\item The ideal $I$ fails the $\mathrm{SLP}$ at the range  $2$ in degree $d-2$.
\item The derivation bundle $\mathcal{D}_0(Z)$ is non-balanced with splitting type $(d-s,d+s+n)$, with $s\geq 1$.
\end{enumerate}
Moreover, if $n$ is even the following third condition is equivalent to the previous two:
\begin{enumerate}[3.]
\item
The derivation bundle $\mathcal{D}_0(Z)$ is unstable with splitting type $(d-s,d+s+n)$, with $s\geq 1$.
\end{enumerate}
\end{prop}
\begin{proof}
It is enough to argue as in \cite[Proposition 7.2]{DIV}.  The ideal $I$ fails the $\mathrm{SLP}$ at the range  $2$ in degree $d-2$ if and only if there exists a curve of degree $d$ through $Z$ and with multiplicity $d-1$ at a general point $P$. By Lemma \ref{linksd}, this condition is equivalent to asking that $\mathcal{D}_0(Z)$ has splitting type $(a,b)$  with $a\leq d-1$, so we can write $a=d-s$ with $s \geq 1$. As the number of points in a line arrangement is always $a+b+1$ we get $b=2d+n-a=d+s+n$ and the difference $|b-a|=2s\geq 2$, so $\mathcal{D}_0(Z)$ has to be non-balanced.
The last part comes from the  equivalence between unstability and the non-balanced condition $|b-a|\geq 2$ that holds when the first Chern class $a+b$ is even.
\end{proof}
Now, the corollary in \cite{DIV} becomes the following.
\begin{prop}\label{prop_arr}
 Let $\mathcal{A}=\{ l_1, \ldots, l_{a+b+1}\}$ be a free line arrangement with exponents $(a,b)$ such that  $a\le b $,  $b-a\ge 2$, let $Z=\{l_1^{\vee},\ldots, l_{a+b+1}^{\vee}\}$ be the corresponding set of points in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^{2\vee}$. For every integer $d$ such that $a+1\leq d\leq \lceil \frac{a+b}2\rceil$\footnote{By $\lceil x\rceil$ we denote the minimum integer greater or equal than $x$}, if the ideals defined below are minimally generated in degree $d$, then
  \begin{enumerate}
\item  if $a+b$ is even, $I=(l_1^d, \ldots, l_{a+b+1}^d)$ fails the SLP at the range $2$ and degree $d-2$;
\item if $a+b$ is odd, let $P$ be a point in general  position with respect to $Z$, then $I=(l_1^{d}, \ldots, l_{a+b+1}^{d},{P^\vee}^{d})$ fails the SLP at the range $2$ and degree  $d-2$.
    \end{enumerate}
\end{prop}
Now, we have to explain how we recognized that in \cite{DIV} the hypothesis on the minimality of generators has been forgotten. We thank the authors of \cite{CHMN16} for the following example.
\begin{ex}\label{CHMN_ex}
Let $\mathcal A=xyz(x+z)(x+2z)\prod_{j=1}^{12}(y+jz)$. It is a free arrangement of splitting type $(3,13)$. The derivation bundle is unstable because it is non-balanced and the first Chern class is even, but the ideal $I=(x^8,y^8,z^8,(x+z)^8,(x+2z)^8,(y+jz)^8|1\leq j\leq 12)$ has the SLP at range $2$.
\end{ex}
\begin{figure}[h!]
    \centering
    \includegraphics[height=4cm]{CHMN.jpg}

    \caption{Dual points of the lines in Example \ref{CHMN_ex}}
  \end{figure}
 Actually, by investigating on the geometric meaning of $0$-syzygies in artinian ideals generated by powers of linear forms, we recognized that they are equivalent to the exisistence of a suitable number of aligned points in $Z$. So examples analogous to Example \ref{CHMN_ex} are the only possibility in order to get $0$-syzygies. In order to state precisely and prove this result,  we recall the following one by Ellia and Peskine \cite[Proposition pag. 112]{EP} related to the numerical character of $Z$.
\begin{defi}\label{defi}Let $S=\mathbb K[x_0,x_1]$ and $Z$ be a $0-$dimensional scheme in the projective plane $\mathbb P^2$. The \em numerical character \em\ of $Z$ is the sequence $(n_0,\ldots,n_{s-1})$ with $n_0\geq \cdots \geq n_{s-1}$ such that
$$0\rightarrow \oplus_{i=0}^{s-1} S(-n_i) \rightarrow \oplus_{i=0}^{s-1} S(-i) \rightarrow \frac{\mathbb K[x_0,x_1,x_2]}{I_Z}\rightarrow 0$$
is a minimal resolution and
\begin{enumerate}
\item $s$ is the minimal degree of a curve containing $Z$;
    \item $n_i \geq s$, for each $i=0,\ldots,s-1$;
        \item\label{huno} Defined $(a)_+= \max\{a,0\}$, $h^1(\mathscr I_Z(n))=\deg(Z)-H_Z(n)=\sum_{0}^{s-1}(n_i-n-1)_+-\sum_{i=0}^{s-1}(i-n-1)_+$, in particular $\deg(Z) = \sum_{i=0}^{s-1}(n_i-i)$.
\end{enumerate}
\end{defi}
\begin{lem}\label{lemma}
Let $Z\subset \mathbb P^2$ a set of points with numerical character $(n_i)_{i=0,\ldots,s-1}$. If $t$ is an integer such that $n_{t-1}>n_t+1$, then there exists a curve $C$ of degree $t$ such that the points of $Z$ contained in $C$ form a set with numerical character $(n_0,\ldots,n_{t-1})$.
\end{lem}
 \begin{thm}\label{th_aligned}
Let $l_1,\ldots,l_{2d+1}$ be linear forms in $\mathbb K[x_0,x_1,x_2]$. Then $$\dim_\mathbb K (l_1^d,\ldots,l_{2d+1}^d)_d<2d+1$$ if and only if in the set $Z=\{l_1^\vee,\ldots,l_{2d+1}^\vee\}\subset {\check{\mathbb P}} \newcommand{\G}{\mathbf G^2}^*$ there are at least $d+2$ aligned points .
\end{thm}
\begin{proof}In the Veronese map $v_d$, aligned points  $l_1^\vee,\ldots,l_r^\vee\in L$ go to points $l_1^d,\ldots,l_r^d\in v_d(L)=C_d$, where $C_d \subset \mathbb P^d$ is the rational normal curve of degree $d$; so $d+1$ of them are linearly independent, while $d+2$ are dependent. So if there are $d+2$ aligned points we get $\dim_\mathbb K (l_1^d,\ldots,l_{2d+1}^d)_d<2d+1$.
 \\
 Conversely, let $\dim_\mathbb K (l_1^d,\ldots,l_{2d+1}^d)_d<2d+1$. By Emsalem-Iarrobino result (\cite{EI}, here Theorem \ref{EI}) we get
 $$
 \dim_{\mathbb K} {I_Z}_d=\dim_{\mathbb K} {\frac R {(l_1^d,\ldots,l_{2d+1}^d)}}_d.
 $$
   So the Hilbert function of $Z$ in degree $d$ is $H_Z(d)=\dim_{\mathbb K}(l_1^d,\ldots,l_{2d+1}^d)_d <2d+1$ and, by \eqref{huno} in Definition \ref{defi}, $h^1(\mathscr I_Z(d))=deg(Z)-H_Z(d)>2d+1-(2d+1)= 0$. With a simple calculation, we deduce that $n_0\geq d+2$. If $n_1\geq d+1$ we would have $2d+2>2d+1=\deg(Z)\geq n_0+n_1-1\geq d+2+d+1-1=2d+2$ and this is a contradiction. So $n_1<n_0-1$ and we can apply Lemma \ref{lemma} with $t=1$ in order to get a line containing a subset $Z'\subset Z$ consisting of $\deg Z'=n_0\geq d+2$ points.
 \end{proof}
So the hypothesis of non-existence of $0$-syzygies is equivalent to the condition that there are at most $d+1$ aligned points in $Z$. In Example \ref{CHMN_ex} there are too many aligned points, actually $14$ that are exactly $14=d+1+s$. Precisely, for $d=8$ in order to avoid linear dependence among $l_1^8,\ldots, l_r^8$, we can have at most $r=9$ aligned points in the dual set  $l_1^\vee,\ldots, l_r^\vee$. But there are $14$ aligned points so there are 5 more, corresponding to 14-9=5 independent $0$-syzygies.

\par With this interpretation of $0$-syzygies we can restate \cite[Proposition 7.2]{DIV} and its corollaries in a nicer geometric way.
\begin{prop}\label{prop_bundle2}
\label{th52}
 Let $I\subset R=\C[x,y,z]$ be an  artinian ideal generated by $\mathbf{2d+1}$  polynomials $l_1^d, \ldots, l_{2d+1}^d$ where $l_i$ are distinct linear forms in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^2$.
Let  $Z=\{l_1^{\vee},\ldots, l_{2d+1}^{\vee}\}$ be the corresponding set of points in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^{2\vee}$. If \textbf{there are no more than $d+1$ aligned points in $Z$}, then the following conditions are equivalent:
\begin{enumerate}
\item The ideal $I$ fails the $\mathrm{SLP}$ at the range  $2$ in degree $d-2$.
\item The derivation bundle $\mathcal{D}_0(Z)$ is unstable with splitting type $(d-s,d+s)$, with $s\geq 1$.
\end{enumerate}
\end{prop}
We recall also Terao's conjecture.
\begin{defi}
Let $\mathcal A$ be a line arrangement, then the combinatorics of $\mathcal A $ is the intersection lattice with reverse order.
\end{defi}\begin{conj}(Terao)
Let $\mathcal A$ be a free arrangement and $\mathcal A'$ an arrangement with the same combinatorics as $\mathcal A$. Then $\mathcal A'$ is free, too (i.e. the freeness is a combinatorial property).
\end{conj}
This conjecture for line arrangements has been proved up to $12$ lines (\cite{FV}).
 In \cite{DIV}, where freeness and unstability are related with failing SLP, an equivalent conjecture
was given in terms of SLP, that here we complete with the missed hypothesis on syzygies.
\begin{conj}\cite{DIV}\label{conj_terao}
Let $\ell_1\cdots \ell_{2b+1}$ and $h_1\cdots h_{2b+1}$ be two arrangements with the same combinatorics and such that the dual sets of points have at most $b+1$ aligned points.
If $I=({{\ell}_1}^b,\ldots,{{\ell}_{2b+1}}^b)$ has SLP at range 2 in degree $b-2$ then also $J=(h_1^b,\ldots,h_{2b+1}^b) $ has SLP at range $2$ in degree $b-2$.
\end{conj}
Now, we end this note with a remark that links Example \ref{CHMN_ex} with Theorem \ref{thgen}.
If there is no $0$-syzygy, $N^- =0$ and the existence of a suitable singular curve is enough to the failure of SLP. Moreover, this hypothesis on $0$-syzygies is necessary, otherwise the unstability is not enough to get an ideal that fails SLP: in Example \ref{CHMN_ex} $H_{R/I}(8)=33=45-(17-5)$ that means that there are $s=5$ independent $0$-syzygies and the ideal has SLP and, moreover, the difference between Hilbert functions is $H_{R/I}(8)-H_{R/I}(6)=-5$. From this we got the idea to generalize Theorem \ref{th1bis} when there are $s$ syzygies in degree $i$ and no syzygy in degree $i-k$ just by replacing the integer $N(r,i,k,d)$ defined in \cite{DIV} with $N(r,i,k,d,s)$ introduced in Definition \ref{def_Ns}. In Example \ref{CHMN_ex} we get $N^-_s=5$ and the existence of singular curves is expected (in the sense of \cite{CHMN16}) and is not enough to the failure of SLP, by applying Theorem \ref{thgen}.
\\ We recall here the definition of unexpected curve.
\begin{defi}(\cite[Definition 2.1]{CHMN16})
We say that a reduced finite set of points $Z\subset \mathbb P^2$ admits an unexpected
curve if there is an integer $j > 0$ such that, for a general point $P$, $jP$ fails to impose the
expected number of conditions on the linear system of curves of degree $j + 1$ containing $Z$.
That is, $Z$ admits an unexpected curve of degree $j + 1$ if
$$
 h^0((\mathcal I_Z \otimes {\mathcal I_P}^j )(j + 1)) > \max \left\{h^0(\mathcal I_Z(j + 1)) -\left( \begin{array}{c}
j + 1\\
2
\end{array}\right)
; 0\right\}.
$$
\end{defi}
A very simple calculation shows that for $k=2,i=0, d=j+1$ we have $$N^-_s=\max \left\{h^0(\mathcal I_Z(j + 1)) -\left( \begin{array}{c}
j + 1\\
2
\end{array}\right)
; 0\right\},$$ so that we have the following special case of \cite[Theorem 6.5]{CHMN16}.
\begin{cor}\label{cor_unexp}
Let $I=(l_1^d, \ldots, l_{2d+1}^d)\subset R=\C[x,y,z]$ be an  artinian ideal generated by $2d+1$ distinct powers of linear forms in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^2$.
Let  $Z=\{l_1^{\vee},\ldots, l_{2d+1}^{\vee}\}$ be the corresponding set of points in $\check{\mathbb P}} \newcommand{\G}{\mathbf G^{2\vee}$.  If there are no more than $d+1$ aligned points in $Z$, then the following are equivalent:
\begin{enumerate}[1.]
\item $Z$ has an unexpected curve of degree $d$;
\item $I$ fails the SLP in range $2$ and degree $d-2$;
\item the variety $\pi_{I_d}(v_d(\check{\mathbb P}} \newcommand{\G}{\mathbf G^2))$ satisfies at least one non trivial Laplace equation of order $d-2$.
\end{enumerate}
\end{cor}
\section*{Acknowledgement} We thank Jean Vall\`{e}s for his valuable comments, David Cook II, Brian Harbourne, Juan Migliore and Uwe Nagel for their remarks on \cite{DIV} and the referee for her/his careful and precise reading and her/his suggestions.
\section*{References}

redpajama_set_name: RedPajamaArXiv


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Tag Archives: Ms.
July 11, 2011 · 1:43 pm
On the Table, 7/11
Some of these items I've been meaning to share for months and am just cleaning out my favorites folders. Sorry for their belatedness.
Good Magazine – "The Most Sustainable Sushi Restaurant in America" – Alex Goldmark
According to a 2011 report from the U.N's Food and Agriculture Organization, "32 percent of world fish stocks are estimated to be overexploited, depleted or recovering and need to be urgently rebuilt."
With conscious consumers starting to take note, Kristofor Lofgren, owner of the Portland restaurant Bamboo Sushi, says his strategy is to overwhelm them with proof that his seafood is actually harvested properly, because, frankly, sushi often isn't. And even when it's not from an illegal vessel or an overfished area, there's still massive waste: As much as half of all fish caught never even make it to the table…. [Read on]
Read more: Looking for sustainable sushi in your neighborhood? Use these sites to score some earth-friendly sushi.
Fish2Fork.com which rates restaurants that serve fish for quality and their sustainability efforts.
The Green Restaurant Association at DineGreen.com which is a national non-profit organization that provides a convenient and cost-effective way for restaurants, manufacturers, distributors, and consumers to become more environmentally responsible.
Live Science – "Why Faces of Other Races Look Alike" – Remy Melina
The well-documented "other-race effect" finds that people are less likely to remember a face from a racial group different from their own. Northwestern University researchers set out to determine what causes this rift in perception and memory by using electroencephalogram (EEG) recordings, which measure brain activity, while participants viewed photos of various faces.
The researchers found that brain activity increases in the very first 200 to 250 milliseconds when seeing both same-race and other-race faces…. [Read on]
Time – "They All Look the Same: How Racism Works Neurologically," by John Cloud on a similar study out of the University of Glasgow which notes that humans are "remarkably skilled at facial recognition: we can differentiate family members and friends from strangers in far less than a second."
Check out The Hapa Project – a book and museum exhibit organized by artist Kip Fulbeck which explores the intersections of race, ethnicity, and identity. When asked, "What are you?" how people of mixed race answer? It's pretty neat.
And your clip for this "On the Table" is of two fabulous female comedians – Tina Fey and Ellen DeGeneres:
It's kind of glitchy. AND some of the good stuff is in the second half, which is here! And part 3 is here!
Check out Tina Fey's new memoir Bossypants about her life so far as a mother and female comedy writer.
Ms. Magazine's Natalie Wilson provides some commentary of the interview here. For example, Wilson states: "While Fey focuses on feminist issues relating to sexism in the workplace and in politics, DeGeneres is more of a comedic lesbian activist. While she has noted she doesn't wish to be an activist or spokesperson for the LGBTQ community, she often publicly denounces heteronormativity and uses her show as a platform to promote LGBTQ rights…." [Read on]
Filed under On the Table
Tagged as Ellen Degeneres, ethnicity, feminism, food, gender, motherhood, Ms., race, Sushi, Sustainable food, The Hapa Project, Tina Fey
Good News for Same-Sex Nuptials and Unions
February 23, 2011 by Stephanie Hallett at Ms. Magazine
In a move eagerly anticipated by gay rights advocates since his election in 2008, President Obama has ordered the Justice Department to cease its defense of the Defense of Marriage Act (DOMA)–bringing federal marriage equality one step closer to reality.
The 1996 act defines marriage as a legal union between a man and woman and allows the federal government to deny recognition to same-sex marriages performed in states where the unions are legal. That means legally married same-sex couples are denied the federal benefits afforded to heterosexual married couples, such as Social Security death benefits and veteran benefits. In directing the Justice Department to stop defending DOMA, constitutional challenges to the law can be brought to court with a hope of succeeding.
University of Pennsylvania law professor Tobias Wolff told The New York Times:
The Justice Department and the president have taken the position on behalf of the United States government that discrimination against gay and lesbian people in all cases is presumptively unconstitutional.
It's the first time the United States government has ever embraced that position, and if the courts agree it will help to eradicate all of the various forms of discrimination that gay and lesbian people suffer around the country.
In another victory for queer couples, same-sex civil unions will soon be allowed in Hawaii. The civil union bill was successful in the state senate today and the governor has vowed to sign it.
The state's relationship with gay civil unions has been tumultuous; in July, a similar bill nearly passed but was vetoed by then-Gov. Linda Lingle. Today, Gov. Neil Abercrombie said, "For me, this bill represents equal rights for all the people of Hawaii."
Though gay rights groups applaud Hawaii's decision, they will continue to push for full marriage equality in that state.
Says Evan Wolfson, founder and president of Freedom to Marry:
Loving and committed same-sex couples have the freedom to marry in 12 countries on four continents–as well as in five states and the District of Columbia here at home–and the sky has not fallen and the sun continues to rise every morning.
While a welcome step, civil union is no substitute for the full measure of respect, clarity, security, responsibilities and protection of marriage itself. States that have created civil union as a means of both giving and withholding–providing legal protections while withholding the freedom to marry and all its meaning–have found that civil union falls far short of marriage with all its tangible and intangible significance in our lives.
Clearly gay rights are taking precedence at the federal level. Last year, a repeal of DADT. This year, an end to DOMA. Upwards, toward an end to discrimination!
Tagged as civil unions, DOMA, Gay Rights, marriage, Ms., New York Times, Obama
Sex and Sexism and "Skins"
Sex, Sexism, and "Skins" by Mia Fontaine at the Ms. blog
MTV, you've come a long way baby.
In six short years you've gone from pimpin' rides to pimpin' girls, starting with the recent premiere of Skins, a remake of the hit British program by the same name. Immediately labeled "the most dangerous show for teens," by the Parents Television Counsel and lambasted for gratuitous sex and drug use, what was seen as most controversial was the employment of underage actors. Given that the show includes implied fellatio and sexual assault, simulated masturbation and half-naked teens having sex, MTV potentially violated child-pornography laws.
Without minimizing the above accusations, what shocked me, however–and should shock everyone–was the show's flagrant and unaddressed sexism. And I'm not talking garden-variety sexism, I'm talking a show that's glaringly sexist in truly alarming ways.
Episode One, for instance, centers around Tony's efforts to get his best friend, the virgin Stanley, laid. His brilliant plan? Borderline date rape. In MTV teen lingo, "get some girl 'recaucusly spliffed. In her confused state she comes to believe how–momentarily of course–you're [Stanley] attractive and then … she bangs your brains out!" For the lucky girl, Tony's girlfriend Michelle nominates Cadie, recently released from a psych ward and described on the Skins' website as "the most dysfunctional girl ever to attend a high school."
Not that Cadie minds. Using sex like wampum, she accepts the plan for Stanley to, "dope me into outer space and then bang my brains out." Is it me, or does this smack of prostitution? Sleeping with someone for drugs cuts the cash out of the equation but not the principle. And if Cadie plays the prostitute, Tony and Michelle play her pimps. Worse, because Michelle is another girl, MTV promotes sexism by all genders. By treating Cadie in a stereotypically male way–a sex object–the message is sent to girls to objectify other girls for male gratification.
Perhaps Michelle's willingness to devalue Cadie as a human being shouldn't surprise us, given her utter subservience to Tony. Despite his belittling nicknaming for her–Nips, because of her "funny nipples"–she continues to see him, and when she does weakly protest the name he patronizingly tells her to "get used to it kid." As if someone appointed him both nipple expert and sage, able to predict a lifetime of nipple shame. Forget worrying over breast size: Now girls are being taught to scrutinize their nipples as well. (It's worth noting that the Skins website describes Michelle as gorgeous and clever. Note to self: MTV defines clever as someone who sleeps with a partner who continuously degrades her).
Another example of the show commodifying young women and encouraging them to use sex to curry favors comes in the second episode, in which the character Tea is asked by her father to go on a date with the son of a prospective business partner.
What is this, the Middle Ages? Aren't we beyond children-as-chattel eras in which daughters do their father's bidding? True, he asks her twice if she's comfortable with it, and reminds her she doesn't have to fool around with the boy (albeit saying "have to" implies the possibility for it). Nonetheless, he tells her not to mention the date to her mother, possibly because Mom would have found it problematic that he used their daughter like a pawn to facilitate a business transaction.
I doubt many people took note of this, however, because Tea's father is the only remotely sympathetic parent on the show. Tony and Stanley's fathers are crass and irate, and the show's mothers, aside from serving food and babysitting, are without real roles or voices. Had Tea's father been a jerk, his request might have raised eyebrows; instead, his affability disguised the fact that, like Cadie being used for Stanley's sexual gain, Tea was used for Daddy's professional gain.
If MTV's looking for edgy, edgy can be done responsibly and respectably. Pierced, tattooed and chain-smoking, The Girl With The Dragon Tattoo's Lisbeth Salander is as edgy and sexual as they come, yet she's subservient to no one and stands up for her beliefs. Given Lisbeth's propensity for justice, I'd love to see what she'd do to the brains behind a show like Skins.
If you want to see this show, be quick about it: Skins might not be around for long. The New York Post reported that it's in danger of cancellation because of low ratings and fleeing advertisers. I just hope the fleeing viewers are as disturbed about the sexism as the underage sex.
Tagged as drugs, media, Ms., MTV, sex, sexism, sexualization of young girls, Skins
Precious Swan – Ms. Magazine
"Precious Swan" by Janell Hobson
[Spoilers included.] There is a painful scene in the critically acclaimed 2009 movie Precious: Based on the Novel Push by Sapphire, which shows our titular heroine (played by Oscar-nominated Gabourey Sidibe) gazing into a mirror and imagining a slender white girl gazing back at her. In the mind of Precious, a 16-year-old poor, black, obese survivor of abuse, she longs to be someone who is her polar opposite–a figure who is beloved and far removed from the sexual abuse that she suffers.
So imagine, then, watching a more recently acclaimed film–Darren Aronofsky's Black Swan, starring Golden Globe winner Natalie Portman–and finding yet another young woman, Nina Sayers, horrified by her own image. Unlike the obese and impoverished Precious Jones, though, Nina is the epitome of beauty and white femininity: an anorexic-thin ballerina, whose chance to shine in the spotlight is nonetheless threatened by darker forces.
Despite their obvious racial, class, physical, and geographic differences (Precious resides in Harlem; Nina lives on the Upper West Side), both women dwell in gloomy apartments heavy with the air of maternal oppression. These heroines could actually be each other's double–a theme dramatically explored in Black Swan. See, both Precious and Nina suffer from mother-daughter sex abuse, a topic that few have discussed publicly even though both films received widespread attention when they opened in theaters. How much of our reluctance has to do not only with the subject matter but also with how each film chooses to tell their stories? How do these stories, and our own responses to them, reflect racial constructions around motherhood and family dysfunction?
When Precious opened last year to critical acclaim and big box office, the main debates surrounding the film concerned the demonization of Precious's mother Mary (played by Oscar-winning Mo'Nique). Black audiences were quick to point out that Mary was a stereotypical distortion of black motherhood, while others chose to view the story through a sociological lens. We could point to the family's impoverished conditions as the cause for its dysfunction, as well as to the sexist treatment both Precious and Mary received from the absentee father, who took off from the household after twice impregnating Precious and giving her HIV, thus sparking Mary's envy and resentment toward her own daughter. This is the pathological black family writ large.
But poverty is just part of the problem, as we witness a similar dysfunctional family in Black Swan, in which Nina and her mother Erica (played by Barbara Hershey) live in a relatively spacious and middle-class apartment in New York City. In this story, there is no mention of the absentee father, which might suggest that single-mother-headed households are viewed as deviant across racial and class lines. However, black dysfunction is often reduced to sociology, while white dysfunction is reduced to psychology. It's not surprising that most conversations about Black Swan focus on how "crazy" and "psychotic" Nina became towards the movie's end, while few have named the source for her mental decline. In other words, few recognize Nina as a victim of incest, the way we immediately recognize Precious's victimization.
Granted, it's rather difficult to not talk about Mary as a sexual predator when director Lee Daniels left little to the imagination (depicting Mary in bed pleasuring herself before calling for Precious to "come take care of Mama"), while Aronofsky overwhelmed Black Swan with surrealism, Swan Lake allegories and ambiguous shots. As a result, the abuse that Nina suffers can only be hinted at (such as depicting Erica calling out to her daughter, "Sweetie, are you ready for me?," right after we witness Nina exploring ways to keep her mother out of her room). Even here, the racial differences in the heroines are highlighted. Precious is expected to serve as the sexual aggressor while Nina is expected to be passive as she awaits her mother in bed.
Moreover, there is a certain ease in blatantly depicting a black mother as monstrous and depraved, while the pathology of the white mother, whose interaction with her daughter is one of suffocating love rather than outright hatred, can only be suggestive and viewed as "creepy" (as in, she is definitely "not the norm," even if Aronofsky invokes ethnic stereotypes of "pushy Jewish mothers").
Even more telling in the conversations we are having about Black Swan is the fetishistic treatment of white women's bodies, which distracts us from some of the deeper themes explored. For instance, the infamous lesbian sex scene, which had audiences buzzing even before the movie debuted, is a cover for the sex abuse Nina experiences. It's important to note that, after this scene (which is later revealed to be an Ecstasy-induced fantasy), Nina's attitude towards her mother changes from fear and resentment to outright disgust, leaving us to question if she had been having sex not with Lily (played by Mila Kunis)–another dancer whom she views as a rival–but with her own mother. It is this disgust that leads to her mental deterioration, her rage (which I interpret as a "healthy" response to abuse) and her eventual embrace of the passionate and sensuous Black Swan role that she must master in her debut performance as the Swan Queen.
Because Nina's abuse remains an unspoken and haunting presence in the film, she cannot find healing. And this is where the narratives in Precious and Black Swan diverge. Even as they tell similar stories of how some daughters suffer unspeakable abuse at the hands of their mothers, we tend to romanticize the story of the black survivor and the white victim. If Precious can find salvation and liberation in an alternative school, where she learns to write and encounters self-empowered women such as her teacher and other students like herself, Nina's education leads to the exact opposite: She can only find self-destruction in art, similarly abusive authority figures (such as her ballet instructor) and female rivalry versus feminist support and friendship. Lily comes closest to being a friend, except Nina's paranoid fantasies, which confuse Lily with her mother, turn her into a threat.
While I realize Aronofsky would rather eschew feminist liberation for his nihilistic fixation on death and madness (a recurring theme throughout his movies), I prefer my own imagined alternate ending for Nina. In my story, she survives her deep wounds, seeks psychiatric treatment and eventually joins a New York City survivors group. There, she might encounter a Precious–both women automatically dismissing each other at first as being too different from themselves. Then they share their stories and realize how easily they could be each other's double.
Tagged as Black Swan, class, feminism, film, gender, mothering, Ms., Precious, race
Club 907 Dancers Freed! : Ms Magazine Blog
Club 907 Dancers Freed! : Ms Magazine Blog.
All of the 80-plus women who were arrested during the November raid of Club 907 in downtown Los Angeles have been released. Club 907 is a "hostess club" where patrons pay for women's companionship (e.g., dancing, delivering drinks). Officers from the L.A. police department (LAPD) raided the club on suspicion of prostitution on November 5, but ended up arresting 78 (of 81) dancers on documentation charges. Although some of the women are still facing immigration court proceedings, this is a major victory for the Coalition for Humane Immigration Rights in Los Angeles (CHIRLA) and the Raids Response Network of the American Immigration Lawyers Association.
The Club 907 raid generated a flurry of criticism. According to the LAPD, the action was initiated because police witnessed illicit sexual activity at the club during a routine inspection. But CHIRLA and others claimed that it constituted an immigration raid in violation of S.B. 40, which prohibits the LAPD from initiating "police action with the objective of discovering the alien status of a person." The vast majority of those arrested were handed over to Immigration and Customs Enforcement (ICE). Others expressed concern that the working conditions may have constituted indentured servitude or human trafficking, and that the LAPD arrested the victims of, and witnesses to, criminal activities at the club.
Speaking of the women's release, Xiomara Corpeño, CHIRLA's organizing director, said:
This is a great day for the almost 80 women who have worked so hard to have their pending immigration cases reconsidered. These women were detained by the LAPD for a crime they were never charged with but ended up nonetheless spending almost two months behind bars. A great injustice was done unto these immigrant workers who are not criminals but working moms and clearly victims of exploitation by their employer. Six out of 10 women were moms and their children missed them deeply during Christmas and New Year's.
The raid on Club 907 generated more than 50 national news stories. The success in this case is a testament to the effectiveness of tireless legal advocacy and intensive media attention in holding public officials accountable.
Tagged as Club 907, ethnicity, feminism, gender, human trafficking, immigration, Ms., prostitution, race, sex trafficking, sex workers
December 27, 2010 · 12:55 pm
How Women Are Faring in the Workplace
Click here to read Catherine Traywick's article explaining the chart at Ms. Magazine.
Tagged as feminism, gender, labor, Ms., work
Ms. Magazine's "No Comment" and Jean Kielborne's "Killing Us Softly"
Every year since Ms. Magazine hit newsstands as part of the roaring Women's Liberation Movement in 1972, the magazine has always included a "No Comment" section on its last page. This page featured advertisements, submitted by readers, that were insulting or degrading to women, but always in the vein of political or social action. Ms. Magazine: "Some make us roll our eyes. Others inspire us to write letters or to boycott products."
And you know what? Ads over the past forty years haven't gotten all that better. For example, here's a classy product for suitcase stickers from July 2010:
Here's what Stephanie Hallett from the Ms. Magazine blog had to say about this:
Identifying your ubiquitous black suitcase on a baggage carousel can be challenging, it's true. But is it really challenging enough to warrant this violent suitcase sticker from thecheeky.com? We think not. Canadian entrepreneur Colin Hart, who runs thecheeky.com, said the stickers are meant to personalize and spice up your travel bags. His collection of large stickers features old leather luggage torn open to reveal illicit contents. What's "inside" the bags? Stacks of cash, bags of cocaine, sex toys–and a bound-and-gagged flight attendant.
Jean Kielborne in her film series "Killing Us Softly" (now in it's fourth edition) provides an in-depth examination of the sexual objectification and degradation of women in advertisements. Watch a snippet from her latest, "Killing Us Softly 4" below:
So what do we do with this? Get involved! One privilege of living in a capitalistic consumer-driven country is the power to put our money where our mouth is. You can submit your images to Ms. Magazine via letterstotheeditor@msmagazine.com, join this Flickr "No Comment" group or start one in another online community. Check out some of the "No Comment" archives for inspiration here.
I'll begin with Urban Outfitters. Can their models get any younger or any skinnier? Or paler?
Tagged as advertisements, consumerism, feminism, gender, Jean Kielborne, Killing us Softly, media, Ms., No Comment, sexism, Urban Outfitters, women's rights
Nude/Sexualized Women in Yoga Ads
Recently Ms. Magazine's blog had an interesting post on the sexualization of women in yoga advertisements, entitled, "Yoga's Feminist Awakening."
So do a Google search for "yoga" and you're going to find a lotta pictures of women. Some naked. Many in bounded, contorted, sexualized positions. Don't believe me?
Hmm, something tells me that more than yoga and goodwill is being sold here.
After seeing the following ads which sell socks (how sexy can you get?),
Yoga Journal Co-founder Judith Hansen Lasater responded:
"I'm concerned about ads that have stimulated both confusion and sadness in me about where the magazine is now and where it is headed. I am confused because I do not understand how photos of naked or half-naked women are connected with the sale of practice products for asana, an important part of yoga. These pictures do not teach the viewer about yoga practice or themselves. They aren't even about the celebration of the beauty of the human body or the beauty of the poses, which I support. These ads are just about selling a product. This approach is something I thought belonged (unfortunately) to the larger culture, but not in Yoga Journal."
And many people have responded to the hoopla with a resounding "who-gives-a-shit." Which is sad.
In the words of Monica Shores via Ms. Magazine: "Looks like some otherwise enlightened yogis could do with a course in Women's Studies 101."
Tagged as Christy Turlington, feminism, gender, Judith Hansen Lasater, Ms., sex, sexual-objectification, toesox, women, yoga, Yoga Journal
"Madonna: Plantation Mistress or Soul Sister?" – bell hooks
In reading this blog post from Ms. Magazine entitled "Why I miss bell hooks" which mentions her essay, "Madonna: Plantation Mistress or Soul Sister?" I decided to share it with all of you. Like Gina Ulysse, I too miss hooks, although I only recently became familiar with her in my graduate career. She was a real, incredibly intelligent soul.
So why this essay? Because it's pretty fun. Madonna's a little outdated, I know, but just try and tell me you don't see 90% of this in Lady Gaga. Her large gay male fan base? Her blond hair? Her controversial "Alejandro" video (just look at whiteness and homosexuality in that one)? Madonna's "emotional cripples" to Lady Gaga's "little monsters"? Anyone? Bueller?
Excerpt from bell hooks, 'Black Looks: Race and Representation.' Read the entire essay here.
No black woman I talked to declared that she wanted to "be Madonna."
Yet we have only to look at the number of black women entertainers/stars (Tina Turner, Aretha Franklin, Donna Summer, Vanessa Williams, Yo-Yo, etc.) who gain greater crossover recognition when they demonstrate that, like Madonna, they too, have a healthy dose of "blonde ambition." Clearly their careers have been influenced by Madonna's choices and strategies. For masses of black women, the political reality that underlies Madonna's and our recognition that this is a society where "blondes" not only "have more fun" but where they are more likely to succeed in any endeavor is white supremacy and racism. We cannot see Madonna's change in hair color as being merely a question of aesthetic choice. I agree with Julie Burchill in her critical work Girls on Film, when she reminds us: "What does it say about racial purity that the best blondes have all been brunettes (Harlow, Monroe, Bardot)? I think it says that we are not as white as we think. I think it says that Pure is a Bore." I also know that it is the expressed desire of the nonblonde Other for those characteristics that are seen as the quintessential markers of racial aesthetic superiority that perpetuate and uphold white supremacy. In this sense Madonna has much in common with the masses of black women who suffer from internalized racism and are forever terrorized by a standard of beauty they feel they can never truly embody.
Like many black women who have stood outside the culture's fascination with the blonde beauty and who have only been able to reach it through imitation and artifice, Madonna often recalls that she was a working-class white girl who saw herself as ugly, as outside the mainstream beauty standard. And indeed what some of us like about her is the way she deconstructs the myth of "natural" white girl beauty by exposing the extent to which it can be and is usually artificially constructed and maintained. She mocks the conventional racist-defined beauty ideal even as she rigorously strives to embody it. Given her obsession with exposing the reality that the ideal female beauty in this society can be attained by artifice and social construction, it should come as no surprise that many of her fans are gay men, and that the majority of nonwhite men, particularly black men, are among that group. Jennie Livingston's film Paris Is Burning suggests that many black gay men, especially queens/divas, are as equally driven as Madonna by "blonde ambition." Madonna never lets her audience forget that whatever "look" she acquires is attained by hard work–"it ain't natural." And as Burchill comments in her chapter "Homosexual Girls": I have a friend who drives a cab and looks like a Marlboro Man but at night is the second best Jean Harlow I have ever seen. He summed up the kind of film star he adores, brutally and brilliantly, when he said, "I like actresses who look as if they've spent hours putting themselves together–and even then they don't look right." Certainly no one, not even die-hard Madonna fans, ever insists that her beauty is not attained by skillful artifice.
And indeed, a major point of the documentary film Truth or Dare: In Bed With Madonna was to demonstrate the amount of work that goes into the construction of her image. Yet when the chips are down, the image Madonna most exploits is that of the quintessential "white girl." To maintain that image she must always position herself as an outsider in relation to black culture. It is that position of outsider that enables her to colonize and appropriate black experience for her own opportunistic ends even as she attempts to mask her acts of racist aggression as affirmation. And no other group sees that as clearly as black females in this society. For we have always known that the socially constructed image of innocent white womanhood relies on the continued production of the racist/sexist sexual myth that black women are not innocent and never can be. Since we are coded always as "fallen" women in the racist cultural iconography we can never, as can Madonna, publicly "work" the image of ourselves as innocent female daring to be bad. Mainstream culture always reads the black female body as sign of sexual experience. In part, many black women who are disgusted by Madonna's flaunting of sexual experience are enraged because the very image of sexual agency that she is able to project and affirm with material gain has been the stick this society has used to justify its continued beating and assault on the black female body.
The vast majority of black women in the United States, more concerned with projecting images of respectability than with the idea of female sexual agency and transgression, do not often feel we have the "freedom" to act in rebellious ways in regards to sexuality without being punished. We have only to contrast the life story of Tina Tumer with that of Madonna to see the different connotations "wild" sexual agency has when it is asserted by a black female. Being represented publicly as an active sexual being has only recently enabled Turner to gain control over her life and career. For years the public image of aggressive sexual agency Turner projected belied the degree to which she was sexually abused and exploited privately. She was also materially exploited. Madonna's career could not be all that it is if there were no Tina Turner and yet, unlike her cohort Sandra Bernhard, Madonna never articulates the cultural debt she owes black females.
In her most recent appropriations of blackness, Madonna almost always imitates phallic black masculinity. Although I read many articles which talked about her appropriating male codes, no critic seems to have noticed her emphasis on black male experience. In his Playboy profile, "Playgirl of the Western World," Michael Kelly describes Madonna's crotch grabbing as "an eloquent visual put-down of male phallic pride." He points out that she worked with choreographer Vince Paterson to perfect the gesture. Even though Kelly tells readers that Madonna was consciously imitating Michael Jackson, he does not contextualize his interpretation of the gesture to include this act of appropriation from black male culture. And in that specific context the groin grabbing gesture is an assertion of pride and phallic domination that usually takes place in an all-male context. Madonna's imitation of this gesture could just as easily be read as an expression of envy.
Throughout [many] of her autobiographical interviews runs a thread of expressed desire to possess the power she perceives men have. Madonna may hate the phallus, but she longs to possess its power. She is always first and foremost in competition with men to see who has the biggest penis. She longs to assert phallic power, and like every other group in this white supremacist society, she clearly sees black men as embodying a quality of maleness that eludes white men. Hence they are often the group of men she most seeks to imitate, taunting white males with her own version of"black masculinity." When it comes to entertainment rivals, Madonna clearly perceives black male stars like Prince and Michael Jackson to be the standard against which she must measure herself and that she ultimately hopes to transcend.
Fascinated yet envious of black style, Madonna appropriates black culture in ways that mock and undermine, making her presentation one that upstages. This is most evident in the video "Like a Prayer." Though I read numerous articles that discussed public outrage at this video, none focused on the issue of race. No article called attention to the fact that Madonna flaunts her sexual agency by suggesting that she is breaking the ties that bind her as a white girl to white patriarchy, and establishing ties with black men. She, however, and not black men, does the choosing. The message is directed at white men. It suggests that they only labeled black men rapists for fear that white girls would choose black partners over them. Cultural critics commenting on the video did not seem at all interested in exploring the reasons Madonna chooses a black cultural backdrop for this ~video, i.e., black church and religious experience. Clearly, it was this backdrop that added to the video's controversy.
In her commentary in the Washington Post, "Madonna: Yuppie Goddess," Brooke Masters writes: "Most descriptions of the controversial video focus on its Catholic imagery: Madonna kisses a black saint, and develops Christ-like markings on her hands. However, the video is also a feminist fairy tale. Sleeping Beauty and Snow White waited for their princes to come along, Madonna finds her own man and wakes him up." Notice that this writer completely overlooks the issue of race and gender. That Madonna's chosen prince was a black man is in part what made the representation potentially shocking and provocative to a white supremacist audience. Yet her attempt to exploit and transgress traditional racial taboos was rarely commented on. Instead critics concentrated on whether or not she was violating taboos regarding religion and representation.
In the United States, Catholicism is most often seen as a religion that has [few] or no black followers and Madonna's video certainly perpetuates this stereotype with its juxtaposition of images of black nonCatholic representations with the image of the black saint. Given the importance of religious experience and liberation theology in black life, Madonna's use of this imagery seemed particularly offensive. For she made black characters act in complicity with her as she aggressively flaunted her critique of Catholic manners, her attack on organized religion. Yet, no black voices that I know of came forward in print calling attention to the fact that the realm of the sacred that is mocked in this film is black religious experience, or that this appropriative "use" of that experience was offensive to many black folk. Looking at the video with a group of students in my class on the politics of sexuality where we critically analyze the way race and representations of blackness are used to sell products, we discussed the way in which black people in the video are caricatures reflecting stereotypes. They appear grotesque. The only role black females have in this video is to catch (i.e., rescue) the "angelic" Madonna when she is "falling." This is just a contemporary casting of the black female as Mammy. Made to serve as supportive backdrop for Madonna's drama, black characters in "Like a Prayer" remind one of those early Hollywood depictions of singing black slaves in the great plantation movies or those Shirley Temple films where Bojangles was trotted out to dance with Miss Shirley and spice up her act. Audiences were not supposed to be enamored of Bojangles, they were supposed to see just what a special little old white girl Shirley really was. In her own way Madonna is a modern day Shirley Temple. Certainly her expressed affinity with black culture enhances her value.
Eager to see the documentary Truth ar Dare because it promised to focus on Madonna's transgressive sexual persona, which I find interesting, I was angered by her visual representations of her domination over not white men (certainly not over Warren Beatty or Alek Keshishian), but people of color and white working-class women. I was too angered by this to appreciate other aspects of the film I might have enjoyed. In Truth or Dare Madonna clearly revealed that she can only think of exerting power along very traditional, white supremacist, capitalistic, patriarchal lines. That she made people who were dependent on her for their immediate livelihood submit to her will was neither charming nor seductive to me or the other black folks that I spoke with who saw the film. We thought it tragically ironic that Madonna would choose as her dance partner a black male with dyed blonde hair. Perhaps had he appeared less like a white-identified black male consumed by "blonde ambition" he might have upstaged her. Instead he was positioned as a mirror, into which Madonna and her audience could look and see only a reflection of herself and the worship of "whiteness" she embodies– that white supremacist culture wants everyone to embody. Madonna used her power to ensure that he and the other nonwhite women and men who worked for her, as well as some of the white subordinates, would all serve as the backdrop to her white-girl-makes-good-drama. Joking about the film with other black folks, we commented that Madonna must have searched long and hard to find a black female that was not a good dancer, one who would not deflect attention away from her. And it is telling that when the film directly reflects something other than a positive image of Madonna, the camera highlights the rage this black female dancer was suppressing. It surfaces when the "subordinates" have time off and are "relaxing."
As with most Madonna videos, when critics talk about this film they tend to ignore race. Yet no viewer can look at this film and not think about race and representation without engaging in forms of denial. After choosing a cast of characters from marginalized groups–nonwhite folks, heterosexual and gay, and gay white folks–Madonna publicly describes them as "emotional cripples." And of course in the context of the film this description seems borne out by the way they allow her to dominate, exploit, and humiliate them. Those Madonna fans who are determined to see her as politically progressive might ask themselves why it is she completely endorses those racist/sexist/classist stereotypes that almost always attempt to portray marginalized groups as "defective" Let's face it, by doing this, Madonna is not breaking with any white supremacist, patriarchal status quo; she is endorsing and perpetuating it.
Some of us do not find it hip or cute for Madonna to brag that she has a "fascistic side," a side well documented in the film. Well, we did not see any of her cute little fascism in action when it was Warren Beatty calling her out in the film. No, there the image of Madonna was the little woman who grins and bears it. No, her "somebody's got to be in charge side," as she names it, was most expressed in her interaction with those representatives from marginalized groups who are most often victimized by the powerful. Why is it there is little or no discussion of Madonna as racist or sexist in her relation to other women? Would audiences be charmed by some rich white male entertainer telling us he must "play father" and oversee the actions of the less powerful, especially women and men of color? So why did so many people find it cute when Madonna asserted that she dominates the interracial casts of gay and heterosexual folks in her film because they are crippled and she "like[s] to play mother" No, this was not a display of feminist power, this was the same old phallic nonsense with white pussy at the center. And many of us watching were not simply unmoved–we were outraged.
Perhaps it is a sign of a collective feeling of powerlessness that many black, nonwhite, and white viewers of this film who were disturbed by the display of racism, sexism, and heterosexism (yes, it's possible to hire gay people, support AIDS projects, and still be biased in the direction of phallic patriarchal heterosexuality) in Truth or Dare have said so little. Sometimes it is difficult to find words to make a critique when we find ourselves attracted by some aspect of a performer's act and disturbed by others, or when a performer shows more interest in promoting progressive social causes than is customary. We may see that performer as above critique. Or we may feel our critique will in no way intervene on the worship of them as a cultural icon. To say nothing, however, is to be complicit with the very forces of domination that make "blonde ambition" necessary to Madonna's success.
Tragically, all that is transgressive and potentially empowering to feminist women and men about Madonna's work may be undermined by all that it contains that is reactionary and in no way unconventional or new. It is often the conservative elements in her work converging with the status quo that have the most powerful impact. For example: Given the rampant homophobia in this society and the concomitant heterosexist voyeuristic obsession with gay life-styles, to what extent does Madonna progressively seek to challenge this if she insists on primarily representing gays as in some way emotionally handicapped or defective? Or when Madonna responds to the critique that she exploits gay men by cavalierly stating: "What does exploitation mean? . . . In a revolution, some people have to get hurt. To get people to change, you have to turn the table over. Some dishes get broken." I can only say this doesn't sound like liberation to me. Perhaps when Madonna explores those memories of her white working-class childhood in a troubled family in a way that enables her to understand intimately the politics of exploitation, domination, and submission, she will have a deeper connection with oppositional black culture. If and when this radical critical self-interrogation takes place, she will have the power to create new and different cultural productions, work that will be truly transgressive–acts of resistance that transform rather than simply seduce.
Tagged as bell hooks, gender, homosexuality, Lady Gaga, Like a Prayer, Madonna, Michael Jackson, Ms., music, pop, Prince, race
Feminist Hulk
By Michelle Kort, Ms.
Ever wanted to know what the feminist Hulk would Twitter? Because it's hilarious.
Here's a bio of Feminist Hulk written by Audrey Bilger:
He's big. He's green. His favorite activity is smashing patriarchy and all forms of oppression. He's Feminist Hulk, and since he first burst onto the Twitter scene less than a month ago he's gathered more than 10,000 followers, who gleefully re-Tweet his 140-character commentaries on gender, feminism and his own personal superhero, feminist theorist Judith Butler. Tweeting in all-caps, this size-XXXXXXL superhero fights for social justice and breaks down the gender binary–all the while looking "smashing" in purple shorts with a big smile on his face.
Here's a peek. Sign up for Tweets here.
HULK FEMINISM ABOUT DISSOLVING HIERARCHIES OF GENDER BINARY, NOT SIMPLY REVERSING THEIR TERMS. ALSO, ABOUT SMASH. AND COOKIES. 9:38 AM Jul 7th via web
HULK STOP FOR QUICK SNACK WHILE SMASHING GENDER BINARY. ACCIDENTALLY SPILL HUMMUS ON TINY PURPLE SHORTS. 12:10 PM Jul 1st via web
HULK USE MALE PRONOUN. HULK SMASH MALE HETEROSEXIST PRIVILEGE. NOT CONTRADICTION, GO TOGETHER LIKE CAKE & ICE CREAM. 1:43 PM Jun 28th via web
Amazingly enough, Feminist Hulk did an interview with Ms. Magazine. Here's a portion, but check out the entire thing here.
Ms.: When did you first discover your feminist superpowers?
feministhulk (to J): WHICH CAME FIRST, THE JUDITH BUTLER OR THE RADIATION?
J: Definitely Butler.
feministhulk: IN BEGINNING, HULK SMASH FOR LOVE OF SMASH. LATER, HULK REALIZE CRAVING FOR SMASH CAUSED BY HEGEMONIC FORCES WHICH DISCONNECTED HULK FROM SELF. HULK QUESTION SYSTEMS OF PRIVILEGE. SOON HULK SMASH WITH GREATER PURPOSE. CULTURAL MINDFULNESS GIVE HULK SUPERPOWERS OF ANTI-PATRIARCHAL SMASH!
J: Like a lot of others, I was given a reductive definition of feminism growing up. I later realized how many kinds of feminism there are, which got me excited about the kind of feminist I wanted to be.
Ms.: Feminist theory often gets a bad rap for being too hard to follow and intimidating. Are you trying to translate theory for a general audience?
feministhulk: HULK LOVE POLYSYLLABIC WORDS. HULK NUZZLE BIG GREEN FACE INTO RICHLY WRITTEN FEMINIST TEXTS. ALSO, THIS JUST HOW HULK TALK.
Ms.: We can't all be superheroes like you. What recommendations do you have for those who might want to nurture their inner Feminist Hulk?"
J: I try not only to pursue change when injustice hits me personally, but also to consider the ways I might be benefiting from privilege without realizing it. Interrogate the places where you are angry, and also interrogate the places where you are comfortable. Be willing to learn new things about your relationship to the world. Be brave enough to keep questioning, keep moving, keep kicking hegemonic ass!
feministhulk: PATRIARCHY NOT WORK ALONE. SEXISM, RACISM, CLASSISM AND HOMOPHOBIA REINFORCE EACH OTHER. HULK SAY TO FOLLOWERS: SMASH ON MULTIPLE FRONTS. AND, JUST IN CASE YOU STUMBLE, BRING EXTRA PAIR OF PURPLE SHORTS.
Tagged as feminism, Feminist Hulk, gender, Gender Trouble, Hulk, Judith Butler, Ms., patriarchy, Twitter

redpajama_set_name: RedPajamaCommonCrawl