3 Afterthought; deleted 16 characters in body; added 39 characters in body

In the case of differential geometry everything reduces to vector spaces. Let $x \in X$. Then at any point $(x, y) \in X \times X$ $$T_{(x, y)} X \times X = T_x X \oplus T_y X .$$ Using this identification the tangent to the diagonal at a point $(x, x)$ is the subspace of $T_{(x, x)} X \times X$ given by $$T_{(x, x)} (\Delta) = \lbrace (\xi, \xi ) \mid \xi \in T_x X \rbrace \subset T_{(x, x)} X \times X.$$ On the other hand the normal is the quotient $$(T_{(x, x)} X \times X) / T_{(x, x)} \Delta$$ and we can identify this with $T_x X$ in at least two slightly different ways. Either $$\iota_1 \colon \xi \mapsto (\xi , - \xi) + T_{(x, x)} \Delta$$ or $$\iota_2 \colon \xi \mapsto (-\xi , \xi) + T_{(x, x)} \Delta .$$

We can of course also identify $T_x X$ and $T_{(x, x)} \Delta$ by $\xi \mapsto (\xi, \xi)$.

I guess one explanation for the two identifications of the normal bundle is that there is involution $\tau \colon X \times X \to X \times X$ given by $\tau(x, y) = (y, x)$ which fixes the diagonal pointwise and hence acts trivially on the tangent space to the diagonal. As a result it descends to an action on the normal bundle which interchanges the two identifications $\iota_1$ and $\iota_2$, that is $\tau \circ \iota_1 = \iota_2$

2 changed a symbol

In the case of differential geometry everything reduces to vector spaces. Let $x \in X$. Then at any point $(x, y) \in X \times X$ $$T_{(x, y)} X \times X = T_x X \times oplus T_y X .$$ Using this identification the tangent to the diagonal at a point $(x, x)$ is the subspace of $T_{(x, x)} X \times X$ given by $$T_{(x, x)} (\Delta) = \lbrace (\xi, \xi ) \mid \xi \in T_x X \rbrace \subset T_{(x, x)} X \times X.$$ On the other hand the normal is the quotient $$(T_{(x, x)} X \times X) / T_{(x, x)} \Delta$$ and we can identify this with $T_x X$ in at least two slightly different ways. Either $$\xi \mapsto (\xi , - \xi) + T_{(x, x)} \Delta$$ or $$\xi \mapsto (-\xi , \xi) + T_{(x, x)} \Delta .$$

We can of course also identify $T_x X$ and $T_{(x, x)} \Delta$ by $\xi \mapsto (\xi, \xi)$.

1

In the case of differential geometry everything reduces to vector spaces. Let $x \in X$. Then at any point $(x, y) \in X \times X$ $$T_{(x, y)} X \times X = T_x X \times T_y X .$$ Using this identification the tangent to the diagonal at a point $(x, x)$ is the subspace of $T_{(x, x)} X \times X$ given by $$T_{(x, x)} (\Delta) = \lbrace (\xi, \xi ) \mid \xi \in T_x X \rbrace \subset T_{(x, x)} X \times X.$$ On the other hand the normal is the quotient $$(T_{(x, x)} X \times X) / T_{(x, x)} \Delta$$ and we can identify this with $T_x X$ in at least two slightly different ways. Either $$\xi \mapsto (\xi , - \xi) + T_{(x, x)} \Delta$$ or $$\xi \mapsto (-\xi , \xi) + T_{(x, x)} \Delta .$$

We can of course also identify $T_x X$ and $T_{(x, x)} \Delta$ by $\xi \mapsto (\xi, \xi)$.