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If $L$ is any line bundle over a complex variety $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.

This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial. In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.

In some cases no square root exists. Some general results are:

1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).

2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different squere roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.

3. If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial unramified double covers of $X$.

The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).

A readable introduction to spinor bundles is provided in the book of MOORE "Lectures on Seiberg-Witten invariants".

If $L$ is any line bundle over a compex manifold $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.

This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial. In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.

In some cases no square root exists. Some general results are:

1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).

2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different square roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.

3. If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial unramified double covers of $X$.

The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).

A readable introduction to spinor bundles is provided in the book of MOORE "Lectures on Seiberg-Witten invariants".

If $L$ is any line bundle over a complex variety $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.

This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial. In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.

In some cases no square root exists. Some general results are:

1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).

2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different squere roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.

3. If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial unramified double covers of $X$.

The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).

If $L$ is any line bundle over a complex variety $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.

This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial. In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.

In some cases no square root exists. Some general results are:

1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).

2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different squere roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.

3. If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial unramified double covers of $X$.

The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).

A readable introduction to spinor bundles is provided in the book of MOORE "Lectures on Seiberg-Witten invariants".

If $L$ is any line bundle over a complex variety $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.

This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial. In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.

In some cases no square root exists. Some general results are:

1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).

2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different squere roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.

3. If $L$ is effective, that is $H^0(X, L) \neq 0$ and $Z \subset X$ is the zero locus of a holomorphic section of $L$, then the existence of a squere root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$.

The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).

If $L$ is any line bundle over a complex variety $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.

This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial. In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.

In some cases no square root exists. Some general results are:

1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).

2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different squere roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.

3. If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial unramified double covers of $X$.

The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).