Let $X\subset{\mathbb P}^d$ be a rational normal curve, $d\gg0$. Then the $k$-th secant variety $Sec^k(X)$ is smooth away from $Sec^{k-1}(X)$, and I wonder what is the singularity of $Sec^k(X)$ at a general point of $Sec^{k-1}(X)$, i.e. the singularity of a 2-dimensional transversal slice. Is it Kleinian by any chance? Does it stabilize as $k$ increases?
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$\begingroup$ I'm posting as a comment due to lack of details. Firstly, $X \subset P^d$ had better be the complete linear embedding of $P^1$ in $P^d$ by polynomials of degree $d$. If we project from a general point on a secant line to the secant variety, we get other singularities. In the case of the rational normal curve, we have a determinantal description of the secant variety as in Harris's first book on AG (p 105). One can proceed inductively to desingularize the secant variety. This was first done in a paper by Bertram on moduli of vector bundles from many years ago. First one blows (continued) $\endgroup$– mehCommented Mar 28, 2014 at 18:18
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$\begingroup$ up the $P^1$ itself. This desingularizes $\text{Sec}^2(C)$ . The fibers are projective spaces, but one still has to use formal functions to show vanishing of the $R^i$. One uses normality of the secant variety and Leray. One continues inductively. $\endgroup$– mehCommented Mar 28, 2014 at 18:21
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$\begingroup$ After explanations by A.Kuznetsov, let me formulate the question more precisely. Let $X={\mathbb P}^1\subset{\mathbb P}^d,\ d=2e$, be a rational normal curve (given in the homogeneous coordinates as $(1,t,t^2,\ldots,t^d)$). It is known that for $k=1,\ldots,e$ the secant variety $Sec^{e-k}X$ is smooth away from $Sec^{e-k-1}X$. I am interested in the singularity of $Sec^{e-k}X$ at a general point of $Sec^{e-k-1}X$. Is it true that there is a transversal slice isomorphic to the Kleinian singularity of type $A_k$ (that is $a^{k+1}=bc$)? $\endgroup$– fnklbergCommented Mar 30, 2014 at 14:01
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