A remark on the genus of curves in ${\mathbf {P}}^4$

Abstract

Let C be an irreducible, reduced, non-degenerate curve, of arithmetic genus g and degree d, in the projective space ${\mathbf {P}}^4$ over the complex field. Assume that C satisfies the following flag condition of type (s, t): C does not lie on any surface of degree $<s$, and on any hypersurface of degree $<t$. Improving previous results, in the present paper we exhibit a Castelnuovo–Halphen type bound for g, under the assumption $s\le t^2-t$ and $d\gg t$. In the range $t^2-2t+3\le s\le t^2-t$, $d\gg t$, we are able to give some information on the extremal curves. They are arithmetically Cohen–Macaulay curves, and lie on a flag like $S\subset F$, where S is a surface of degree s, F a hypersurface of degree t, S is unique, and its general hyperplane section is a space extremal curve, not contained in any surface of degree $<t$. In the case $d\equiv 0$ (modulo s), they are exactly the complete intersections of a surface S as above, with a hypersurface. As a consequence of previous results, we get a bound for the speciality index of a curve satisfying a flag condition.

Appendix

We keep all the notation of Sects. 1 and 2.

(i) The function $\rho =\rho (s,t,\epsilon )$ is defined as follows (see [2, p. 120]¹).

If $\epsilon \ge s-(\beta +1)(\alpha +\beta +2-t)$, divide $s-\epsilon -1=u(\alpha +\beta +2-t)+v$, and put:

$$\begin{aligned} \rho:= & {} \frac{s-1-\epsilon }{s}\left[ \frac{s^2}{2t}+\frac{s}{2}(t-4)-\frac{(t-1-\beta )(1+\beta )(t-1)}{2t}+1\right] +\frac{1+\epsilon }{2s}\left( s-\epsilon +1\right) \\&+{\left( {\begin{array}{c}u+v+1\\ 2\end{array}}\right) }-\frac{1}{2}(\alpha +\beta )(2v+u\alpha +u\beta -u^2)+\frac{1}{2}u(t-1)(t-\beta -3)-1; \end{aligned}$$

if $\epsilon < s-(\beta +1)(\alpha +\beta +2-t)$, divide $\epsilon =u(\alpha +\beta +1)+v$, and put:

$$\begin{aligned} \rho:= & {} \frac{s-1-\epsilon }{s}\left[ \frac{s^2}{2t}+\frac{s}{2}(t-4)-\frac{(t-1-\beta )(1+\beta )(t-1)}{2t}+1\right] +\frac{1+\epsilon }{2s}(s-\epsilon +1)\\&-\frac{1}{2}(\alpha +\beta )(t-u-1)(\alpha +t+u-3)+\frac{1}{2}\beta (t-1)(2\alpha -6)-1. \end{aligned}$$

Similarly, we define $\rho '=\rho (s,\tau ,\epsilon )$ (compare with Sect. 2, (i)).

(ii) The number R appearing in (7) is defined as follows ([1, 5, pp. 91–92, (4) and (4$^\prime$)]).

First, define k and $\delta$ by dividing $\epsilon =kw+\delta$, $0\le \delta <w$, when $\epsilon < (3-w_1)w$. Otherwise, define k and $\delta$ by dividing $\epsilon +2-w_1=k(w+1)+\delta$, $0\le \delta < w+1$. Then we have:

$$\begin{aligned} R:=\frac{1+\epsilon }{2s}(s+1-\epsilon -2\pi )+w(\epsilon -\delta )-k{\left( {\begin{array}{c}w+1\\ 2\end{array}}\right) }+{\left( {\begin{array}{c}\delta \\ 2\end{array}}\right) }. \end{aligned}$$

(iii) Sketch of the proof of (6) We only prove that $|\rho |\le 2t^3$ in the case $\epsilon \ge s-(\beta +1)(\alpha +\beta +2-t)$. The analysis of the case $\epsilon < s-(\beta +1)(\alpha +\beta +2-t)$, and the proof of the estimate $|\rho '|\le 2t^3$, are quite similar, therefore we omit them.

Set:

$$\begin{aligned} H:=H(s,t):=\frac{s^2}{2t}+\frac{s}{2}(t-4)-\frac{(t-1-\beta )(1+\beta )(t-1)}{2t}+1. \end{aligned}$$

(17)

This number is the coefficient of the term $\frac{s-1-\epsilon }{s}$ appearing in the definition of $\rho$. By the way, notice that, if $s>t^2-t$, then H is the Halphen’s bound for the genus of a space curve of degree s, not contained in any surface of degree $<t$ ([9, p. 1], [7, 10.8. Teorema, p. 56]). We also notice we may write:

$$\begin{aligned} G(d,s,t)=\frac{d^2}{2s}+\frac{d}{2s}\left( 2H-2-s\right) +\rho +1. \end{aligned}$$

(18)

Taking into account that $s-1=\alpha t+\beta$, we may rewrite H:

$$\begin{aligned} 2H=\alpha t^2+(\alpha ^2-4\alpha )t+\beta ^2+2\alpha +(2\alpha -1)\beta . \end{aligned}$$

(19)

The function $\alpha \rightarrow \alpha ^2-4\alpha$ is growing for $\alpha \ge 2$. Therefore, when $\alpha \ge 2$, since $0\le \alpha \le t-2$ and $0\le \beta \le t-1$, it follows that:

$$\begin{aligned} 0\le 2H\le & {} (t-2)t^2+((t-2)^2-4(t-2))t+(t-1)^2 \nonumber \\&+2(t-2)+(2(t-2)-1)(t-1) =2\left( t^3-\frac{7}{2}t^2+\frac{5}{2}t+1\right) . \end{aligned}$$

(20)

This inequality holds true also for $\alpha \le 1$. Hence

$$\begin{aligned} 0\le H\le t^3-\frac{7}{2}t^2+\frac{5}{2}t+1. \end{aligned}$$

(21)

Since $\frac{s-1-\epsilon }{s}\ge 0$, $H\ge 0$, and $t-\beta -3\ge -2$, from the definition of $\rho$ we deduce:

$$\begin{aligned} \rho \ge -\frac{1}{2}(\alpha +\beta )(2v+u\alpha +u\beta )-u(t-1)-1. \end{aligned}$$

Taking into account that

$$\begin{aligned} s\le t^2-t,\quad \alpha \le t-2,\quad v\le \alpha +\beta +1-t\le \beta -1\le t-2, \quad u\le \beta \le t-1, \end{aligned}$$

(22)

substituting in a similar manner as in (20), it follows that

$$\begin{aligned} \rho \ge -2t^3+5t^2-\frac{3}{2}t-\frac{7}{2}\ge -2t^3. \end{aligned}$$

(23)

Moreover, since $\frac{s-1-\epsilon }{s}\le 1$,

$$\begin{aligned} \frac{1+\epsilon }{2s}(s-\epsilon +1)\le \frac{1}{2}(s+1)\le \frac{1}{2}(t^2-t+1), \end{aligned}$$

and $u\le \beta$ (which implies that $u\beta -u^2\ge 0$, so $-\frac{1}{2}(\alpha +\beta )(2v+u\alpha +u\beta -u^2)\le 0$), from (21), (22), and the definition of $\rho$, it follows that:

$$\begin{aligned} \rho\le & {} H + \frac{1}{2}(t^2-t+1)+\frac{1}{2}(u+v+1)(u+v)+\frac{1}{2}ut(t-1) \\\le & {} \left( t^3-\frac{7}{2}t^2+\frac{5}{2}t+1\right) + \frac{1}{2}(t^2-t+1)+\frac{1}{2}(2t-2)(2t-3)+\frac{1}{2}t(t-1)^2 \\= & {} \frac{3}{2}t^3-2t^2-\frac{5}{2}t+\frac{9}{2}\le 2t^3. \end{aligned}$$

Combining this estimate with (23), we deduce $|\rho |\le 2t^3$, in the case $s\le t^2-t$ and $\epsilon \ge s-(\beta +1)(\alpha +\beta +2-t)$.

(iv) Proof of (8) Recall that $s-1=2w+w_1$, $0\le w_1\le 1$, and $\pi =w(w-1+w_1)$ (compare with Sect. 2, (iii), and with this “Appendix”, (ii)). Hence we have:

$$\begin{aligned} s+1-\epsilon -2\pi \le s+1-2\pi =\frac{1}{2}(-s^2+6s+w_1^2-2w_1-1)\le \frac{1}{2}(6s-s^2). \end{aligned}$$

Therefore, if $s\ge 6$, then $s+1-\epsilon -2\pi \le 0$. In this case, taking into account that $w\le (s-1)/2$ and that $\delta \le w$, we have:

$$\begin{aligned} R\le & {} w\epsilon +\frac{1}{2}\delta (\delta -1)\le w(s-1)+\frac{1}{2}w(w-1)\\\le & {} \frac{1}{2}(s-1)^2+\frac{1}{2}\frac{s-1}{2}\frac{s-3}{2}\le \frac{5}{8}(s-1)^2\le s^2. \end{aligned}$$

An easy direct computation shows that the inequality $R\le s^2$ holds true also when $3\le s\le 5$. Therefore we have:

$$\begin{aligned} R\le s^2. \end{aligned}$$

(24)

On the other hand we have:

$$\begin{aligned} R\ge \frac{1+\epsilon }{2s}(-\epsilon -2\pi )-w\delta -\frac{1}{2}w(w+1)k. \end{aligned}$$

Hence:

$$\begin{aligned} -R\le & {} \frac{1}{2}(s-1+2\pi )+w\delta +\frac{1}{2}w(w+1)k \nonumber \\= & {} \frac{1}{2}\left( \frac{1}{2}s^2-s+\frac{1}{2}+w_1-\frac{1}{2}w_1^2\right) +w\delta +\frac{1}{2}w(w+1)k. \end{aligned}$$

(25)

When $\epsilon < w(3-w_1)$, then $\delta \le w-1$ and $k\le 2$. Therefore, in this case, from (25) we have:

$$\begin{aligned} -R\le & {} \frac{1}{2}\left( \frac{1}{2}s^2-s+1\right) +w(w-1)+w(w+1)\\= & {} \frac{1}{2}\left( \frac{1}{2}s^2-s+1\right) +2w^2\le \frac{1}{2}\left( \frac{1}{2}s^2-s+1\right) +2\frac{(s-1)^2}{4}= \frac{3}{4}s^2-\frac{3}{2}s+1\le s^2. \end{aligned}$$

When $\epsilon \ge w(3-w_1)$, then $\delta \le w$ and $k\le \frac{2(s+1)}{s}$. From (25) we get:

$$\begin{aligned} -R\le & {} \frac{1}{2}\left( \frac{1}{2}s^2-s+1\right) +w^2+w(w+1)\frac{s+1}{s}\\\le & {} \frac{1}{2}\left( \frac{1}{2}s^2-s+1\right) +\frac{(s-1)^2}{4}+(s^2-1)\frac{s+1}{4s}= \frac{1}{4s}\left( 3s^3-3s^2+2s-1\right) \le s^2. \end{aligned}$$

Combining with (24), we get $|R|\le s^2$.

(v) Proof of Lemma 2.1 Consider the coefficient of $\frac{d}{2}$ in the expression defining G(d, s, t) and $G(d,s,\tau )$ (Sect. 2, (ii)):

$$\begin{aligned}&A:=\frac{s}{t}+t-5-\frac{(t-1-\beta )(1+\beta )(t-1)}{st},\\&A':=\frac{s}{\tau }+\tau -5-\frac{(\tau -1-\beta ')(1+\beta ')(\tau -1)}{s\tau }. \end{aligned}$$

We have:

$$\begin{aligned} G(d,s,\tau )=G(d,s,t)+\frac{d}{2}(A'-A)+(\rho '-\rho ). \end{aligned}$$

(26)

Observe that (compare with (17)):

$$\begin{aligned} H=\frac{s}{2}(A+1)+1, \quad H'=\frac{s}{2}(A'+1)+1, \end{aligned}$$

where $H'=H(s,\tau )$. Hence, by (19) (compare with Sect. 2, (i)), we have:

$$\begin{aligned} A'-A= & {} \frac{2}{s}(H'-H)=\frac{1}{s}\left[ \alpha '\tau ^2+(\alpha '^2-4\alpha ')\tau +\beta '^2 +2\alpha '+(2\alpha '-1)\beta '\right] \\&-\frac{1}{s}\left[ \left( \alpha t^2+(\alpha ^2-4\alpha )t+\beta ^2 +2\alpha +(2\alpha -1)\beta \right) \right] . \end{aligned}$$

Simplifying, we get:

$$\begin{aligned} A'-A=\frac{\alpha x}{s}(\alpha x+\alpha +x-2t+3+2\beta ). \end{aligned}$$

Hence, if $\alpha x=0$, then $A'=A$. When $\alpha >0$ and $x>0$, since $-(t-\beta )<-x(\alpha +1)$, we have:

$$\begin{aligned} \alpha x+\alpha +x-2t+3+2\beta= & {} \alpha x+\alpha +x-2(t-\beta )+3\\\le & {} \alpha x+\alpha +x-2(x(\alpha +1)+1)+3=(1-x)(\alpha +1)\le 0. \end{aligned}$$

If $x=1$, then the number

$$\begin{aligned} \alpha x+\alpha +x-2t+3+2\beta =2\alpha +4-2t+2\beta \end{aligned}$$

vanishes if and only if $\beta =t-\alpha -2$. Summing up, we get: in any case, one has $A'\le A$. Moreover, $A'=A$if and only if either $\alpha =0$or $x=0$or $x=1$and $\beta =t-\alpha -2$, i.e. if and only if either $s\le t$or $s\ge t+1$and $t-\alpha -2\le \beta <t$. In particular, when $A'<A$, then $A-A'\ge \frac{\alpha }{s}\ge \frac{1}{st}$.

We deduce the following.

(1) If $t+1\le s\le t^2-t$ and $\beta <t-\alpha -2$, then $A'<A$. Therefore, from (6) and (26), we deduce that $G(d,s,\tau ) < G(d,s,t)$ for $d>8st^4$. In fact, in this case, we have $\frac{d}{2}(A'-A)+(\rho '-\rho )<0$, because $\frac{2(\rho '-\rho )}{A-A'}\le 2\cdot 4t^3\cdot st$.

(2) If $s\le t^2-t$ and $t-\alpha -2<\beta$, then $A=A'$. Hence, (26) becomes $G(d,s,\tau )=G(d,s,t)+(\rho '-\rho )$. A direct computation, which we omit, shows that, in this case, if either $s-\epsilon -1<\alpha +\beta +2-t$ or $\beta (\alpha +\beta +2-t)\le s-\epsilon -1<(\beta +1)(\alpha +\beta +2-t)$, then $\rho =\rho '$. Hence, we have $G(d,s,\tau )=G(d,s,t)$.

(3) If either $s\le t$ or $t+1 \le s\le t^2-t$ and $t-\alpha -2\le \beta$, then $A=A'$. Therefore, by (6) and (26), we get $G(d,s,\tau )=G(d,s,t)+(\rho '-\rho )\le G(d,s,t)+4t^3$.

This concludes the proof of Lemma 2.1.

(vi) Proof of Lemma 2.2 Consider the coefficient of $\frac{d}{2}$ in the formula (7) defining G:

$$\begin{aligned} A'':=\frac{2\pi -2}{s}-1. \end{aligned}$$

We have:

$$\begin{aligned} G=G(d,s,t)+\frac{d}{2}(A''-A)+(R-\rho -1). \end{aligned}$$

(27)

A direct computation proves that:

$$\begin{aligned} A''-A=\frac{1}{2s}\left[ (\alpha ^2-2\alpha )t^2+(2\alpha \beta +6\alpha -2\alpha ^2)t+ (-4\alpha -\beta ^2-4\alpha \beta +w_1)\right] . \end{aligned}$$

If $\alpha =0$, i.e. $s\le t$, then $s=\beta +1$, and

$$\begin{aligned} A''-A=\frac{1}{2s}(-\beta ^2+w_1). \end{aligned}$$

If $t+1\le s\le 2t-3$, then $\alpha =1$, and we have:

$$\begin{aligned} A''-A=\frac{1}{2s}\left[ -(t-\beta -2)^2+w_1\right] . \end{aligned}$$

In both cases we have $A''-A<-\frac{1}{2s}$. Therefore, from (27) and (8), we deduce that $G < G(d,s,t)$ for $d> 32t^4$, because in this case $\frac{d}{2}(A''-A)+(R-\rho -1)<0$ (in fact: $\frac{2(R-\rho -1)}{A-A''}\le 4s(s^2+2t^3)\le 32t^4$).

This concludes the proof of Lemma 2.2.

Remark 6.1

A similar argument shows that if $2t-2\le s\le 2t$, then $A''=A$, and that if $t>2$ ed $s\ge 2t+1$, then $A''>A$. Moreover, notice that, when $s\ge t+1$ and $t-\alpha -2\le \beta$, it may happen that $G(d,s,\tau )>G(d,s,t)$. For instance, if $s=t^2-2t+6$ and $\epsilon =s-25$, then $\rho '-\rho =2(t+1)$.

(vii) Proof of (12) Since $t^2-2t+3\le s\le t^2-t$, we have

$$\begin{aligned} s-1=(t-2)t+\beta ,\quad 2\le \beta \le t-1. \end{aligned}$$

Inserting into (17), we get:

$$\begin{aligned} H(s,t)= & {} \frac{((t-2)t+\beta +1)^2}{2t}+\frac{(t-2)t+\beta +1}{2}(t-4)-\frac{(t-1-\beta )(1+\beta )(t-1)}{2t}+1\\= & {} t^3-5t^2+(\beta +7)t+\frac{\beta ^2-5\beta }{2}-2\\= & {} \left( t^3-5t^2+(\beta +7)t+\frac{\beta ^2-7\beta }{2}-1\right) +\beta -1= P(s,t)+\beta -1. \end{aligned}$$