Yes, the square root of a line bundle $L$ is a line bundle $L'$ such that $(L')^{\otimes 2} \simeq L.$

I don't know of any general criteria for detecting the existence of a square root. A couple basic observations:

• You certainly need the degree of $L$ to be even for a square root to exist.

• (edited based on David Speyer's comment below) If $U_i$ form an open cover of your manifold, then the transition functions for $L$ are given by maps $g_{ij}:U_i \cap U_j \to GL_1$ satisfying the cocycle condtion. These of course are the same as elements $g_{ij} \in \mathcal{O}(U_i \cap U_j)^{\ast}.$ To have a square root, it would suffice for a square root of the $g_{ij}$ to exist in $\mathcal{O}(U_i \cap U_j)^{\ast},$ as these would form transition functions for the square root of $L.$ As David Speyer notes in the comments, the converse is not true.

In regards to your specific question regarding the canonical line bundle of $\mathbb{P}^1,$ this is easy. The canonical bundle is $\mathcal{O}(-2),$ and its square root is the tautological line bundle $\mathcal{O}(-1).$ Since isomorphism classes of line bundles on $\mathbb{P}^1$ are determined completely by their degree, a line bundle on $\mathbb{P}^1$ has a square root if and only if its degree is even.

Yes, the square root of a line bundle $L$ is a line bundle $L'$ such that $(L')^{\otimes 2} \simeq L.$

I don't know of any general criteria for detecting the existence of a square root. A couple basic observations:

• You certainly need the degree of $L$ to be even for a square root to exist.

(second edit: This second bullet point is still incorrect. See David Speyer's comments and Tyler Lawson's answer for a correct formulation.)

• (edited based on David Speyer's comment below) If $U_i$ form an open cover of your manifold, then the transition functions for $L$ are given by maps $g_{ij}:U_i \cap U_j \to GL_1$ satisfying the cocycle condtion. These of course are the same as elements $g_{ij} \in \mathcal{O}(U_i \cap U_j)^{\ast}.$ To have a square root, it would suffice for a square root of the $g_{ij}$ to exist in $\mathcal{O}(U_i \cap U_j)^{\ast},$ as these would form transition functions for the square root of $L.$ As David Speyer notes in the comments, the converse is not true.

In regards to your specific question regarding the canonical line bundle of $\mathbb{P}^1,$ this is easy. The canonical bundle is $\mathcal{O}(-2),$ and its square root is the tautological line bundle $\mathcal{O}(-1).$ Since isomorphism classes of line bundles on $\mathbb{P}^1$ are determined completely by their degree, a line bundle on $\mathbb{P}^1$ has a square root if and only if its degree is even.

Yes, the square root of a line bundle $L$ is a line bundle $L'$ such that $(L')^{\otimes 2} \simeq L.$

I don't know of any general criteria for detecting the existence of a square root. A couple basic observations:

• You certainly need the degree of $L$ to be even for a square root to exist.

• If $U_i$ form an open cover of your manifold, then the transition functions for $L$ are given by maps $g_{ij}:U_i \cap U_j \to GL_1$ satisfying the cocycle condtion. These of course are the same as elements $g_{ij} \in \mathcal{O}(U_i \cap U_j)^{\ast}.$ To have a square root, you would need a square root of the $g_{ij}$ to exist in $\mathcal{O}(U_i \cap U_j)^{\ast},$ as these would form transition functions for the square root of $L.$

In regards to your specific question regarding the canonical line bundle of $\mathbb{P}^1,$ this is easy. The canonical bundle is $\mathcal{O}(-2),$ and its square root is the tautological line bundle $\mathcal{O}(-1).$ Since isomorphism classes of line bundles on $\mathbb{P}^1$ are determined completely by their degree, a line bundle on $\mathbb{P}^1$ has a square root if and only if its degree is even.

Yes, the square root of a line bundle $L$ is a line bundle $L'$ such that $(L')^{\otimes 2} \simeq L.$

I don't know of any general criteria for detecting the existence of a square root. A couple basic observations:

• You certainly need the degree of $L$ to be even for a square root to exist.

• (edited based on David Speyer's comment below) If $U_i$ form an open cover of your manifold, then the transition functions for $L$ are given by maps $g_{ij}:U_i \cap U_j \to GL_1$ satisfying the cocycle condtion. These of course are the same as elements $g_{ij} \in \mathcal{O}(U_i \cap U_j)^{\ast}.$ To have a square root, it would suffice for a square root of the $g_{ij}$ to exist in $\mathcal{O}(U_i \cap U_j)^{\ast},$ as these would form transition functions for the square root of $L.$ As David Speyer notes in the comments, the converse is not true.

In regards to your specific question regarding the canonical line bundle of $\mathbb{P}^1,$ this is easy. The canonical bundle is $\mathcal{O}(-2),$ and its square root is the tautological line bundle $\mathcal{O}(-1).$ Since isomorphism classes of line bundles on $\mathbb{P}^1$ are determined completely by their degree, a line bundle on $\mathbb{P}^1$ has a square root if and only if its degree is even.