Let $p$ be a line of the vector space $K^{n+1}$, and let $H$ be a hyperplane of $K^{n+1}$ such that $p\subseteq H$. We may also interpretate $H$ as a line $q\subseteq(K^{n+1})^{*}$.

Let $\alpha\in\mathrm{Hom}(p,K^{n+1}/p), \beta\in\mathrm{Hom}(H,K^{n+1}/H)$. We can understand $\beta\in\mathrm{Hom}(q,(K^{n+1})^{*}/q)$. Let $v\in p,w\in q$. In p.208 of Harris's 'Algebraic Geometry: A first course' it is said that the condition $$ \langle\alpha(v),w \rangle+\langle v,\beta(w)\rangle=0\quad\text{ for } v\in p,w\in q $$ where $\langle, \rangle$ is the map of the dual pairing, is equivalent to $$ \beta|_{p}\equiv\alpha \text{ }(\mathrm{mod}\text{ } H). $$ I don't really understand why this is true. Any hint would be appreciated.

  • $\begingroup$ I tried proving this by using the identifications which seems natural to me, but I arrive at $\beta|_p \equiv -\alpha \pmod H$. Could you maybe elaborate at how your identifications look like? (I suspect this is where I am missing some sign.) $\endgroup$ – Jendrik Stelzner Jan 17 '16 at 1:12
  • $\begingroup$ @JendrikStelzner Harris is studying the tangent space of an incidence correspondence of the kind $I=\{(p,H)\in\mathbb{P}^{n}\times (\mathbb{P}^{n})^{*}:p\in H\}$ (We use the same name por a projective subspace and its associated vector space in $K^{n+1}$). At first he says is that $T_{(p,H)}I=\{(\alpha,\beta)\in\mathrm{Hom}(p,K^{n+1}/p)\times\mathrm{Hom}(H,K^{n+1}/H):\beta|_{p}\equiv\alpha\text{ }(\mathrm{mod}\text{ }H)\}$, but that, considering $H$ as a point $q\in(\mathbb{P}^{n})^{*}=\mathbb{P}((K^{n+1})^{*})$, it is equivalent to $\endgroup$ – H. Jackson Jan 17 '16 at 8:36
  • $\begingroup$ @JendrikStelzner $T_{(p,H)}I=\{(\alpha,\beta)\in\mathrm{Hom}(p,K^{n+1}/p)\times\mathrm{Hom}(q,(K^{n+1})^{*}/q:\langle\alpha(v),w \rangle+\langle v,\beta(w)\rangle=0\text{ for } v\in p,w\in q\}$ $\endgroup$ – H. Jackson Jan 17 '16 at 8:37
  • $\begingroup$ @JendrikStelzner I am a bit stuck. I'd appreciate if you could tell me how have you achieved $\beta|_{p}\equiv -\alpha\text{ }(\mathrm{mod}\text{ }H)$. Thank you! $\endgroup$ – H. Jackson Jan 19 '16 at 11:37
  • $\begingroup$ While I’m not sure about how to read the original statements in Harris’ book I added my calculations on your question. $\endgroup$ – Jendrik Stelzner Jan 21 '16 at 20:03

This is not a complete answer to the question, but only my try so far, as requested by the question author. In particular I do not arrive at the given answer, but at one having different signs. As I am not sure about the used identifications I will try to make all identifications explicit, albeit at the costs of elegance.

We may also interpret $H$ as a line $q \subseteq (K^{n+1})^*$.

I assume that we are doing so by taking the orthogonal complement of $H$, i.e. $$ q = \{w \in (K^{n+1})^* \mid \text{$w(h) = 0$ for every $h \in H$}\}. $$

Let $\alpha \in \mathrm{Hom}(p,K^{n+1}/p)$ […]

Because $p \subseteq H$ we can identify $q$ with a subspace of $(K^{n+1}/p)^*$ via $$ q \to (K^{n+1}/p)^*, \quad w \mapsto w' \quad\text{where}\quad w'(x+p) := w(x) \quad\text{for all $x \in K^{n+1}$} $$ Similary we can identify $q$ with a subspace of $(K^{n+1}/H)^*$ via $$ q \to (K^{n+1}/H)^*, \quad w \mapsto w'' \quad\text{where}\quad w''(y+H) = w(y) \quad\text{for all $y \in K^{n+1}$}. $$ (This is actually already an isomorphism.)

[…] $\beta \in \mathrm{Hom}(H, K^{n+1}/H)$. We can understand $\beta \in \mathrm{Hom}(q, (K^{n+1})^*/q)$.

I assume we do so in the following way:

We first identify $K^{n+1}/H$ with $q^*$ by using that $$ K^{n+1} \to q^*, \quad x \mapsto (w \mapsto w(x)) $$ is surjective with kernel $H$; the resulting isomorphism is give by $$ K^{n+1}/H \to q^*, \quad x \mapsto (w \mapsto w''(x)). $$ This results in an isomorphism $\mathrm{Hom}(H, K^{n+1}/H) \to \mathrm{Hom}(H, q^*)$, under which $\beta_1 := \beta \in \mathrm{Hom}(H, K^{n+1}/H)$ corresponds to $\beta_2 \in \mathrm{Hom}(H, q^*)$ given by $$ \beta_2(h)(w) = w''(\beta(h)) $$

Now we use the natural ismorphism $$ \Phi \colon \mathrm{Hom}(H, q^*) \to \mathrm{Hom}(q, H^*), \quad \Phi(f)(w)(h) = f(h)(w). $$ Under $\Phi$ the element $\beta_2 \in \mathrm{Hom}(H, q^*)$ corresponds to $\beta_3 \in \mathrm{Hom}(q, H^*)$ given by $$ \beta_3(w)(h) = \Phi(\beta_2)(w)(h) = \beta_2(h)(w) = w''(\beta(h)). $$

Lastly we identify $(K^{n+1})^*/q$ with $H^*$ via the restriction $$ (K^{n+1})^* \to H^*, \quad \phi \mapsto \phi|_H, $$ which is surjective with kernel $q$. Under the induced isomorphism $\mathrm{Hom}(q, (K^{n+1})^*/q) \to \mathrm{Hom}(q, H^*)$ the preimage of $\beta_3 \in \mathrm{Hom}(q, H^*)$ is given by $\beta_4 \in \mathrm{Hom}(q, (K^{n+1})^*/q)$.

[…] the condition $$ \langle \alpha(v), w \rangle + \langle v, \beta(w) \rangle = 0 \quad\text{for every $v \in p$, $w \in q$} $$ where $\langle \cdot, \cdot \rangle$ is the map of the dual pairing […]

Using the above identifications this condition is expressed by $$ \langle \alpha(v), w' \rangle + \langle v, \beta_3(w) \rangle = 0 \quad\text{for every $v \in p$, $w \in q$} $$

Now let $v \in p$, and let $\alpha(v) = x+p$ and $\beta(v) = y+H$ with $x,y \in K^{n+1}$. Then $$ \langle \alpha(v), w' \rangle = w'(\alpha(v)) = w'(x+p) = w(x) $$ and $$ \langle v, \beta(w) \rangle = \beta_3(w)(v) = w''(\beta(v)) = w''(y+H) = w(y). $$ Therefore \begin{align*} &\, \langle \alpha(v), w' \rangle + \langle v, \beta_3(w) \rangle = 0 \quad\text{for every $w \in q$} \\ \iff&\, w(x) + w(y) = 0 \quad\text{for every $w \in q$} \\ \iff&\, \langle x+y, w \rangle = 0 \quad\text{for every $w \in q$} \\ \iff&\, x + y \in H \\ \iff&\, x \equiv -y \pmod{H} \\ \iff&\, \alpha(v) \equiv -\beta(v) \pmod{H}, \end{align*} So we have \begin{align*} &\, \langle \alpha(v), w' \rangle + \langle v, \beta_3(w) \rangle = 0 \quad\text{for every $v \in p$, $w \in q$} \\ \iff&\, \alpha(v) \equiv -\beta(v) \pmod{H} \quad\text{for every $v \in p$} \\ \iff&\, \alpha \equiv -\beta|_p \pmod{H}. \end{align*}

While this is not the desired result I hope that I helps and that I just overlooked a sign somewhere.

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