Let $X$ be a smooth irreducible algebraic curve in $\mathbb{P} V$. The general position lemma states that the points given by general hyperplane section of $X$ are "in general position". I'm considering a similar situation for tangent lines. For each point of the general hyperplane section, the tangent line at that point corresponds to a point in $\mathbb{P} ( \wedge^2 V)$. Is it true that these tensors in $\wedge^2V$ span the whole ambient space $\wedge^2 V$? (Of course, one should assume that the degree of $X$ is bigger than $\dim (\wedge^2 V)$). I suspect if this question is related to (the surjectivity of) the Gauss map.

$\begingroup$ It seems to me that a possible rephrasing of your question would be 'Does the image of the Gauss map span the entire Projective space $P\bigwedge^2(V)$ . Surjectivity of the Gauss map is related to the map being a closed immersion and says nothing about the span of the map. I personally know of no results on this topic. $\endgroup$– mehCommented Feb 28, 2014 at 2:47

$\begingroup$ @aginensky: I agree with your comment, I think the OP is confusing two different notions. Maybe if the curve is both nondegenerate and projectively normal ...? By the way, good to run into you again (if only virtually)! $\endgroup$– Jason StarrCommented Feb 28, 2014 at 18:16

$\begingroup$ @ Jason Thanks. Many time reader, but only occasional commenter. Btw, on a related note, if one considers the 'Gauss map' for secant varieties to curves, I know of Nothing about the image. Do you know any literature on this ? $\endgroup$– mehCommented Mar 1, 2014 at 14:41
1 Answer
No, that is not true. Let $V$ be the $4$dimensional vector space with ordered basis $(\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)$, and let $[X_0,X_1,X_2,X_3]$ be the corresponding homogeneous coordinates on $\mathbb{P}V$. Let $[T_0,T_1]$ be homogeneous coordinates on $\mathbb{P}^1$. For every integer $r\geq 3$, consider the morphism, $$ u_r:\mathbb{P}^1 \to \mathbb{P}V, \ \ [T_0,T_1] \mapsto [T_0^r\mathbf{e}_0 + T_0^{r1}T_1\mathbf{e}_1 + T_0T_1^{r1}\mathbf{e}_2 + T_1^r\mathbf{e}_3].$$ The corresponding Gauss map is $$\widetilde{u}_r: \mathbb{P}^1 \to \mathbb{P}(\bigwedge^2V),$$ $$[T_0,T_1] \mapsto [rT_0^{2r2}\mathbf{e}_0\wedge \mathbf{e}_1 + r(r1)T_0^rT_1^{r2}\mathbf{e}_0\wedge\mathbf{e}_2 + r^2T_0^{r1}T_1^{r1}\mathbf{e}_0\wedge \mathbf{e}_3$$ $$ + r(r2)T_0^{r1}T_1^{r1}\mathbf{e}_1\wedge \mathbf{e}_2 + r(r1)T_0^{r2}T_1^r\mathbf{e}_1\wedge\mathbf{e}_3 + rT_1^{2r2}\mathbf{e}_2\wedge \mathbf{e}_3]. $$ In particular, the coefficients of $\mathbf{e}_0\wedge\mathbf{e}_3$ and $\mathbf{e}_1\wedge\mathbf{e}_2$ are proportionate, i.e., the following "linear form" vanishes on the image curve, $$(r2)X_0\wedge X_3  rX_1\wedge X_2.$$ Notice that $r$ can be arbitrarily positive.

$\begingroup$ Thanks a lot! But I'm considering a curve given by a complete intersection, for which case it can be proven to be tangentially nondegenerate. Therefore, the locus of tangent lines is nondegenerate in $\mathbb{P} (\wedge^2 V)$, unlike the example given in the answer. What can be said under this additional assumption that the given curve is tangentially nondegenerate? $\endgroup$ Commented Mar 1, 2014 at 14:46

$\begingroup$ I just want to clarify the question. I believe you are specifying that $u:C\hookrightarrow \mathbb{P}(V)$ is a smooth, nondegenerate curve of degree $d \geq \binom{n}{2}$ such that the Gauss map, $\widetilde{u}:C\hookrightarrow \mathbb{P}(\bigwedge^2 V)$, is also nondegenerate. Now you ask, for a general choice $H$ of hyperplane in $\mathbb{P}(V)$, does $\widetilde{u}(u^{1}(H))$ span $\mathbb{P}(\bigwedge^2 V)$. Is this your question? $\endgroup$ Commented Mar 1, 2014 at 15:03

$\begingroup$ Yes, that is exactly what I wanted to know. Thanks for clarigying the question. By the way, it was nice for me to learn a simple example you explained above with a linear form vanishing on the Gauss image. $\endgroup$ Commented Mar 1, 2014 at 17:20