How kmalloc() and vmalloc() used in 32bit vs 64bit systems?

Question

Architecture: X86, Linux

I have gone through several threads related to kmalloc() and vmalloc() and I know the basic differences between them but still have some doubts. I need your help to understand the fundamentals.

So As we know that in 32-bit systems the virtual address space is divided between user and kernel. The high 1GB memory addresses are allocated for kernel and the lower 3GB to user space.

Please correct me i am wrong here:

So out of 1GB 896MB is 1:1 mapped (physically contiguous) to the kernel logical address and 128MB is used to access the high memory addresses means accessing the physical(RAM) beyond 896 MB by mapping it into the 128MB using vmalloc()? (lets consider we have 2GB of RAM)
And what would happen if we have only 1GB of physical RAM in point 1. The 896MB is 1:1 mapping between kernel logical address and physical addresses achieved through kmalloc(). Then what would be the use of 128MB, why do we need vmalloc() anymore? So lets say if we have 1GB of RAM which entirely can be accessed through kernel logical address so does that mean that it is possible only when the amount of memory requested by a thread/process can be allocated in physically contiguous addresses. And if there is no physical contiguous memory available, in that case memory allocated will not be physically contiguous, hence we need to use vmalloc() to map it in 128MB of space and access it through there?
What happens when we use 64-bit system? Is it true that both kmalloc() and vmalloc() can have enormous amount of address space and can access entire RAM?

have you read [this document](https://elinux.org/images/b/b0/Introduction_to_Memory_Management_in_Linux.pdf), perhaps it might help — Sabrina Jewson, May 29 '19 at 06:51
I had already gone through that document but still have doubts. — Usr1, May 30 '19 at 09:53

How kmalloc() and vmalloc() used in 32bit vs 64bit systems?

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