In Joe Harris's text book entitled : Algebraic geometry, a first course, page : $ 55 $, we find the following paragraph which i don't really understand :
... As another example, consider again twisted cubics. Any twisted cubic : $ C $ can be written as the image of the map of the form : $$ t \to [ a_{03} t^3 + a_{02} t^2 + a_{01} t + a_{00} , \dots , a_{33} t^3 + a_{32} t^2 + a_{31} t + a_{30} ] $$ where the determinant of the matrix $ (a_{ij} ) $ of coefficients is nonzero, so that we can describe $ C $ by specifying a non singular $ 4 \times 4 $ matrix. Of course, this does not give a bijection between the variety $ U \subset (K)^{16} $ of invertible $ 4 \times 4 $ matrices and the set of twisted cubics, since the curve $ C $ does not determine this expression.
Alternately, recall from theorem $ 1.18 $ that any twisted cubic is determined by six points in general position in $ \mathbb{P}^3 $. Thus, we get a twisted cubic for any point in the open subset $ V \subset ( \mathbb{P}^3 )^6 $ corresponding to configurations in general position, though as in the previous case, this is not a bijection.
Could you explain to me please, in onother simpler way why there is no bijection between the variety $ U \subset (K)^{16} $ of invertible $ 4 \times 4 $ matrices and the set of twisted cubics ? It's not quite clear for me this small text above.
Thanks in advance for your help.
The text does not say that there is no bijection (there are for instance when $K=\Bbb R$ or $\Bbb C$, since both sets are continuously infinite), but that the map described here does not provide a bijection.
The reason why the present map does not give a bijection is that you can multiply all terms $a_{i,j}$ by a fixed non-zero element of $K$, this will give you another matrix but will describe the same twisted cubic.